09243500460  09243500460

Chapter 13: Limits and Derivatives

Exercise 13.1

Q1.

Evaluate \(\lim_{x \rightarrow 3} x + 5\) limit:

Soln:

\(\lim_{x \rightarrow 3} x + 5 = 3 + 5 = 8\)

 

Q2.

Evaluate \(\lim_{x \rightarrow \pi}\left ( x – \frac{22}{7} \right )\)

Soln:

\(\lim_{x \rightarrow \pi} ( x – \frac{22}{7}) =( \pi – \frac{22}{7})\)

 

Q3.

Evaluate \(\lim_{r \rightarrow l} \pi r^{2}\)

Soln:

\(\lim_{r \rightarrow l} \pi r^{2} = \pi (l)^{2} = \pi \)

 

Q4.

Evaluate \(\lim_{x + 4}\frac{4x + 6}{x – 3}\)

Soln:

\(\lim_{x + 4}\frac{4x + 6}{x – 3} = \frac{4 (4) + 6}{4 – 3} = \frac{16 + 6}{1} = \frac{22}{1} = 22\)

 

Q5.

Evaluate \(\lim_{x \rightarrow -1} \frac{x^{12} + x^{7} + 1}{x – 1}\)

Soln:

\(\lim_{x \rightarrow -1} \frac{x^{12} + x^{7} + 1}{x – 1} = \frac{(-1)^{12} + (-1)^{7} + 1}{-1 – 1 } = \frac{1 – 1 + 1}{-2} = -\frac{1}{2}\)

 

Q6.

Evaluate \(\lim_{x \rightarrow 0}\frac{(x + 3)^{5} – 3}{x}\)

Soln:

\(\lim_{x \rightarrow 0}\frac{(x + 3)^{5} – 3}{x}\)

Put x + 3 = y so that \(y \rightarrow 1 \; as \; x \rightarrow 0\)

Accordingly, \(\\ \lim_{x \rightarrow 0}\frac{(x + 3)^{5} – 3}{x} = \lim_{y \rightarrow 1}\frac{y^{5} – 3}{y – 3} \\ = \lim_{y \rightarrow 1}\frac{y^{5} – 3^{5}}{y – 3} \\ = 5.3^{5 – 1} \\ = 405 \\ ∴ \lim_{x \rightarrow 0}\frac{(x + 3)^{5} – 3}{x} = 405\)

 

Q7.

Evaluate \(\lim_{x \rightarrow 2}\frac{3x^{2} – x – 10}{x^{2} – 4}\)

Soln:

At x = 2, the given rational function has the value of 0/ 0

\(\\ \lim_{x \rightarrow 2}\frac{3x^{2} – x – 10}{x^{2} – 4} = \lim_{x \rightarrow 2}\frac{ \left ( x – 2 \right ) \left ( 3x + 5 \right ) }{\left ( x – 2 \right ) \left ( x + 2 \right ) } \\ = \lim_{x \rightarrow 2} \frac{3x + 5}{x + 2} \\ = \frac{3 (2) + 5}{2 + 2} \\ = \frac{11}{4}\)

 

Q8.

Evaluate \(\\ \lim_{x \rightarrow 3}\frac{x^{4} – 81}{2x^{2} – 5x – 3}\)

Soln:

At x = 2, the given rational function has the value of 0/ 0

\(\\ \lim_{x \rightarrow 3}\frac{x^{4} – 81}{2x^{2} – 5x – 3} = \lim_{x \rightarrow 3}\frac{(x- 3)(x + 3)(x^{2} + 9)}{(x – 3)(2x + 1)} \\ = \lim_{x \rightarrow 3}\frac{(x + 3)(x^{2} + 9)}{2x + 1} \\ = \frac{(3 + 3)(3^{2} + 9)}{2(3) + 1} \\ = \frac{6 \times 18}{7} \\ = \frac{108}{7}\)

 

Q9.

Evaluate \(\lim_{x \rightarrow 0}\frac{ax + b}{cx + 1}\)

Soln:

\(\lim_{x \rightarrow 0}\frac{ax + b}{cx + 1} = \frac{a(0) + b}{c(0)+ 1} = b\)

 

Q10.

Evaluate \(\lim_{z \rightarrow 1}\frac{z^{\frac{1}{3}} – 1}{z^{\frac{1}{6}} – 1}\)

Soln:

\(\lim_{z \rightarrow 1}\frac{z^{\frac{1}{3}} – 1}{z^{\frac{1}{6}} – 1}\)

At z = 1, the given function has the value of 0/ 0

Put \(z^{\frac{1}{6}} = x \; so \; that \; z\rightarrow 1 \; as \; x \rightarrow 1 \\ Accordingly, \; \lim_{z \rightarrow 1} \frac{z^{\frac{1}{3}} – 1}{z^{^{\frac{1}{6}}} – 1} = \lim_{x \rightarrow 1} \frac{x^{2} – 1}{x – 1} \\ = \lim_{x \rightarrow 1}\frac{x^{2} – 1^{2}}{x – 1} \\ = 2.1^{2 – 1} \\ = 2 \\ ∴ \lim_{z \rightarrow 1}\frac{z^{\frac{1}{3}} – 1}{z^{\frac{1}{6}} – 1} = 2\)

 

Q11.

Evaluate \(\lim_{x \rightarrow 1}\frac{ax^{2} + bx + c}{cx^{2} + bx + a}, \; a + b + c \neq 0\)

Soln:

\(\\ \lim_{x \rightarrow 1}\frac{ax^{2} + bx + c}{cx^{2} + bx + a} = \frac{a(1)^{2} + b(1) + c}{c(1)^{2} + b(1) + a}\\ = \frac{a + b + c}{a + b + c} \\ = 1\)

 

Q12.

\(\lim_{x \rightarrow 2}\frac{\frac{1}{x} + \frac{1}{2}}{x + 2}\)

Soln:

\(\lim_{x \rightarrow 2}\frac{\frac{1}{x} + \frac{1}{2}}{x + 2}\)

At x = -2, the given function has the value of 0/ 0

\(Now, \; \lim_{x \rightarrow 2} \frac{\frac{1}{x} + \frac{1}{2}}{x + 2} = \lim_{x \rightarrow -2} \frac{\left ( \frac{2 + x}{2x} \right )}{x + 2} \\ = \lim_{x \rightarrow -2}\frac{1}{2x} \\ = \frac{1}{2 (-2) } = \frac{-1}{4}\)

 

Q13.

Evaluate \(\lim_{x \rightarrow 0}\frac{\sin ax}{bx}\)

Soln:

\(\lim_{x \rightarrow 0}\frac{\sin ax}{bx}\)

At x = 0, the given function has the value of 0/ 0

\(\\ Now, \lim_{x \rightarrow 0}\frac{\sin ax}{bx} = \lim_{x \rightarrow 0} \frac{sin ax}{ax} \times \frac{ax}{bx} \\ = \lim_{x \rightarrow 0}\left ( \frac{\sin ax}{ax} \right ) \times \left ( \frac{a}{b} \right ) \\ = \frac{a}{b} \lim_{ax \rightarrow 0}\left ( \frac{\sin ax}{ax} \right ) \\ = \frac{a}{b} \times 1 \\ = \frac{a}{b}\)

 

Q14.

Evaluate \(\lim_{x \rightarrow 0}\frac{\sin ax}{\sin bx}, a, \; b \neq 0\)

Soln:

\(\lim_{x \rightarrow 0}\frac{\sin ax}{\sin bx}, a, \; b \neq 0\)

At x = 0, the given function has the value of 0/ 0

\(\\ Now, \lim_{x \rightarrow 0}\frac{sin ax}{sin bx} = \lim_{x \rightarrow 0} \frac { \left (\frac{ \sin ax}{ax} \right ) \times ax}{\left ( \frac{\sin bx}{bx} \right ) \times bx}\\ = \left ( \frac{a}{b} \right ) \times \frac{\lim_{ax \rightarrow 0}\left ( \frac{\sin ax}{ax} \right )}{\lim_{bx \rightarrow 0} \left ( \frac{\sin bx}{bx} \right )} \\ = \left ( \frac{a}{b} \right ) \times \frac{1}{1} \\ = \frac{a}{b}\)

 

Q15.

Evaluate \(\lim_{x \rightarrow \pi}\frac{\sin (\pi – x)}{\pi (\pi – x)}\)

Soln:

\(\lim_{x \rightarrow \pi}\frac{\sin (\pi – x)}{\pi (\pi – x)}\)

It seems \(x \rightarrow \pi \Rightarrow \left ( \pi – x \right ) \rightarrow 0\)

\(\\ ∴ \lim_{x \rightarrow \pi} \frac{\sin (\pi – x)}{\pi (\pi – x)} = \frac{1}{\pi}\lim_{\left ( z – x \right ) – 1}\frac{\sin(\pi – x)}{(\pi – x)} \\ = \frac{1}{\pi} \times 1 \\ = \frac{1}{\pi}\)

 

Q16.

Evaluate \(\lim_{x \rightarrow 0}\frac{\cos x}{\pi – x}\)

Soln;

\(\lim_{x \rightarrow 0}\frac{\cos x}{\pi – x} = \frac{\cos 0}{\pi – 0} = \frac{1}{\pi}\)

 

Q17.

Evaluate \(\lim_{x \rightarrow 0} \frac{\cos 2x – 1}{\cos x – 1}\)

Soln:

\(\lim_{x \rightarrow 0} \frac{\cos 2x – 1}{\cos x – 1}\)

At x = 0, the given function has the value of 0/ 0

\(\\ Now, \\ \lim_{x \rightarrow 0}\frac{\cos 2x – 1}{\cos x – 1} = \lim_{x \rightarrow 0}\frac{1 – 2 \sin^{2} x – 1}{1 – 2 \sin^{2} \frac{x}{2} – 1} \\ = \lim_{x \rightarrow 0}\frac{\sin^{2}x}{\sin^{2}\frac{x}{2}} = \lim_{x \rightarrow 0}\frac{\left ( \frac{\sin^{2}x}{x^{2}} \right ) \times x^{2}}{\left ( \frac{sin^{2}\frac{x}{2}}{\left ( \frac{x}{2} \right )^{2}} \right )\times \frac{x^{2}}{4}} \\ = 4 \frac{\lim_{x\rightarrow 0}\left ( \frac{\sin^{2}x}{x^{2}} \right )}{\lim_{x \rightarrow 0} \left ( \frac{sin^{2}\frac{x}{2}}{\left ( \frac{x}{2} \right )^{2}} \right )}\) \(\\ = 4 \frac{\left (\lim_{x \rightarrow 0}\frac{\sin x}{x} \right )^{2}}{\left ( \lim_{\frac{x}{2} \rightarrow 0} \frac{sin \frac{x}{2}}{\frac{x}{2}} \right )^{2}} \\ = 4\frac{1^{2}}{1^{2}} \\ = 4\)

 

Q18.

Evaluate \(\lim_{x \rightarrow 0}\frac{ax + x \cos x}{b \sin x}\)

Soln:

\(\lim_{x \rightarrow 0}\frac{ax + x \cos x}{b \sin x}\)

At x = 0, the given function has the value of 0/ 0

\(\\ Now \\ \lim_{x \rightarrow 0}\frac{ax + x \cos x}{b \sin x} = \frac{1}{b} \lim_{x \rightarrow 0}\frac{x(a + \cos x)}{\sin x}\\ = \frac{1}{b} \lim_{x \rightarrow 0} \left ( \frac{x}{\sin x} \right ) \times \lim_{x \rightarrow 0}(a + \cos x) \\ = \frac{1}{b} \times \frac{1}{\left ( \lim_{x \rightarrow 0} \frac{\sin x}{x} \right )} \times \lim_{x \rightarrow 0}\left ( a + \cos x \right ) \\ = \frac{1}{b} \times \left ( a + \cos 0 \right ) \\ = \frac{a + 1}{b}\)

 

Q19.

Evaluate \(\lim_{x \rightarrow 0} x \sec x\)

Soln:

\(\lim_{x \rightarrow 0} x \sec x = \lim_{x \rightarrow 0}\frac{x}{\cos x} = \frac{0}{\cos 0} = \frac{0}{1} = 0\)

 

Q20.

Evaluate \(\lim_{x \rightarrow 0}\frac{\sin ax + bx}{ax + \sin bx} a, \;b \; \; a + b \neq 0\)

Soln:

At x =0, the given function has the value of 0/ 0

\(\\ Now, \\ \lim_{x \rightarrow 0}\frac{\sin ax + bx}{ax + \sin bx} \\ \lim_{x \rightarrow 0}\frac{\left ( \frac{\sin ax}{ax} \right )ax + bx}{ax + bx\left ( \frac{\sin bx}{bx} \right )} \\ =\) \(\\ = \frac{\left ( \lim_{ax \rightarrow 0} \frac{\sin ax}{ax} \right ) \times \lim_{x \rightarrow 0}\left ( ax \right ) + \lim_{x \rightarrow 0} bx}{\lim_{x \rightarrow 0}ax + \lim_{x \rightarrow 0} bx\left ( \lim_{bx \rightarrow 0} \frac{\sin bx}{bx} \right )} \\ = \frac{\lim_{x \rightarrow 0}(ax) + \lim_{x \rightarrow 0}bx}{\lim_{x \rightarrow 0}ax + \lim_{x \rightarrow 0}bx} \\ = \frac{\lim_{x \rightarrow 0}(ax + bx)}{\lim_{x \rightarrow 0}(ax + bx)} \\ = \lim_{x \rightarrow 0}(1) \\ = 1\)

 

Q21.

Evaluate \(\lim_{x \rightarrow 0}(\csc x – \cot x)\)

Soln:

At x =0, the given function has the value of ∞ – ∞

\(\\ Now, \\ \lim_{x \rightarrow 0}(\csc x – \cot x) \\ = \lim_{x \rightarrow 0}\left ( \frac{1}{\sin x} – \frac{\cos x}{\sin x} \right ) \\ = \lim_{x \rightarrow 0}\left ( \frac{1 – \cos x}{\sin x} \right ) \\ = \lim_{x \rightarrow 0}\frac{\left (\frac{1 – \cos x}{\sin x} \right )}{\left (\frac{\sin x}{x} \right )} \\ = \frac{\lim_{x \rightarrow 0}\frac{1 – \cos x}{x}}{\lim_{x \rightarrow 0}\frac{\sin x}{x}} \\ = \frac{0}{1} \\ = 0\)

 

Q22.

\(\lim_{x \rightarrow \frac{\pi}{x}}\frac{\tan 2x}{x – \frac{\pi}{2}}\)

Soln:

\(\lim_{x \rightarrow \frac{\pi}{x}}\frac{\tan 2x}{x – \frac{\pi}{2}}\)

At \( x = \frac{\pi}{2}\), the given function has the value of 0/ 0

\(\\ Now, \; put \; x – \frac{\pi}{2} = y \; so \; that \; x \rightarrow \frac{\pi}{2}, \; y \rightarrow 0 \\ ∴ \lim_{x \rightarrow \frac{\pi}{2}}\frac{\tan2x}{x – \frac{\pi}{2}} = \lim_{y \rightarrow 0}\frac{tan 2\left ( y + \frac{\pi}{2} \right ) }{y} \\ = \lim_{y \rightarrow 0} \frac{tan \left ( \pi + 2y \right ) }{y} \\ =\lim_{y \rightarrow 0} \frac{\tan 2y}{y} \\ = \lim_{y \rightarrow 0}\frac{\sin 2y}{y \cos 2y} \\ = \lim_{y \rightarrow 0}\left ( \frac{\sin 2y}{2y} \times \frac{2}{\cos 2y} \right ) \\ = \left ( \lim_{2y \rightarrow 0} \frac{\sin 2y}{2y} \right ) \times \lim_{y \rightarrow 0}\left ( \frac{2}{\cos 2y} \right ) \\ =1 \times \frac{2}{\cos 0} \\ = 1 \times \frac{2}{1} \\ = 2\)

 

Q23.

Find \(\\ \lim_{x\rightarrow 0} f(x) \; and \; \lim_{x\rightarrow 0} f(x), \\ where f(x) = \left\{\begin{matrix} 2x + 3 & x \leq 0\\ 3(x + 1) & x > 0 \end{matrix}\right.\)

Soln:

Given function \(f(x) =\{\begin{matrix} 2x + 3 & x \leq 0\\ 3(x + 1) & x > 0 \end{matrix}\)

 

\(\lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0}\left [ 2x + 3 \right ] = 2(0) + 3 = 3\)

 

\(\lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} 3(x + 1) = 3(0 + 1) = 3\)

 

\(∴ \lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} f(x) = 3\)

 

\(\lim_{x \rightarrow 1} f(x) = \lim_{x \rightarrow 1} 3(x + 1) = 3(1 + 1) = 6\)

 

\(\lim_{x \rightarrow 1} f(x) = \lim_{x \rightarrow 1} 3(x + 1) = 3(1 + 1) = 6\)

 

\(∴ \lim_{x \rightarrow 1} f(x) = \lim_{x \rightarrow 1} f(x) = \lim_{x \rightarrow 1} f(x)\) = 6

 

Q24.

Find \(\lim_{x \rightarrow 1} f(x), \; where \; f(x) = \left\{\begin{matrix} x^{2} – 1, & x \leq 1 \\ -x^{2} – 1, & x \geq 1 \end{matrix}\right.\)

Soln:

Given function is

\(f(x) = \left\{\begin{matrix} x^{2} – 1, & x \leq 1 \\ -x^{2} – 1, & x \geq 1 \end{matrix}\right.\) \(\lim_{x \rightarrow 1^{-} f(x)} = \lim_{x \rightarrow 1}[x^{2} – 1] = 1^{2} – 1 = 1 – 1 = 0\) \(\lim_{x \rightarrow 1^{+} f(x)} = \lim_{x \rightarrow 1}[-x^{2} – 1] = -1^{2} – 1 = 1 – 1 = 0\)

We observed that \(\lim_{x \rightarrow 1^{-}} f(x) \neq \lim_{x \rightarrow 1^{+}} f(x)\)

Hence, \(\lim_{x \rightarrow 1} f(x)\) doesn’t exist.

 

25.

Evaluate \(\lim_{x \rightarrow 0} f(x), \; where \; f(x) = \left\{\begin{matrix} \frac{\left | x \right |}{x}, & x \neq 0 \\ 0, & x = 0 \end{matrix}\right.\)

Soln:

Given function \(f(x) = \left\{\begin{matrix} \frac{\left | x \right |}{x}, & x \neq 0 \\ 0, & x = 0 \end{matrix}\right.\)

When \( \left | x \right | = -x \)

\(\lim_{x \rightarrow 0 ^{-}} f(x) = \lim_{x \rightarrow 0^{-}}\left [ \frac{\left | x \right |}{x} \right ]\)

= \(\lim_{x \rightarrow 0 \left ( \frac{ -x }{x} \right )}\)

= \(\lim_{x \rightarrow 0 \left ( -1 \right )}\)

= -1

When \( \left | x \right | = x \)

\(\lim_{x \rightarrow 0 ^{+}} f(x) = \lim_{x \rightarrow 0^{+}}\left [ \frac{\left | x \right |}{x} \right ]\)

= \(\lim_{x \rightarrow 0 \left [ \frac{ -x }{x} \right ]}\)

= \(\lim_{x \rightarrow 0 \left ( 1 \right )}\)

= 1

We observe that \(\lim_{x \rightarrow 0^{-}} f(x) \neq \lim_{x \rightarrow 0^{+}} f(x)\)

Hence, \(\lim_{x \rightarrow 0 } f(x) \) doesn’t exist.

 

Q26.

Find \(\lim_{x \rightarrow 0} f(x), \; where \; f(x) = \left\{\begin{matrix} \frac{x}{\left | x \right |}, & x \neq 0 \\ 0, & x = 0 \end{matrix}\right.\)

Soln:

f(x) = \(\{\begin{matrix} \frac{x}{| x |}, & x \neq 0 \\ 0, & x = 0 \end{matrix}\)

When \(\left | x \right | = -x\)

\(\lim_{x \rightarrow 0^{-}}f(x) = \lim_{x \rightarrow 0^{-}}\left [ \frac{x}{\left | x \right |} \right ]\) \( = \lim_{x \rightarrow 0}\left [ \frac{x}{ – x } \right ]\) \( = \lim_{x \rightarrow 0}\left ( – 1 \right )\)

= -1

When \(\left | x \right | = x\)

\(\lim_{x \rightarrow 0^{+}}f(x) = \lim_{x \rightarrow 0^{+}}\left [ \frac{x}{\left | x \right |} \right ]\) \( = \lim_{x \rightarrow 0}\left [ \frac{x}{ x } \right ]\) \( = \lim_{x \rightarrow 0}\left ( 1 \right )\)

= 1

We observe that \(\lim_{x \rightarrow 0^{-}} f(x) \neq \lim_{x \rightarrow 0^{+}} f(x)\)

Hence, \(\lim_{x \rightarrow 0 } f(x) \) doesn’t exist.

 

Q27.

Find \(\lim_{x \rightarrow 5} f(x), \; where \; f(x) = \left | x \right | – 5\)

Soln:

Given function is \( f(x) = \left | x \right | – 5\)

When \( x > 0, \left | x \right | = x \)

\(\lim_{x \rightarrow 5^{-}} f(x) = \lim_{x \rightarrow 5^{-}}[\left | x \right | – 5]\)

= \( \lim_{x \rightarrow 5} \left ( x – 5\right )\)

= 5 – 5

= 0

When \( x > 0, \left | x \right | = x \)

\(\lim_{x \rightarrow 5^{+}} f(x) = \lim_{x \rightarrow 5^{+}}(\left | x \right | – 5)\)

= \( \lim_{x \rightarrow 5} \left ( x – 5\right )\)

= 5 – 5

= 0

\(∴ \lim_{x \rightarrow 5^{-}}f(x) = \lim_{x \rightarrow 5^{+}}f(x) = 0\)

Hence, \(\lim_{x \rightarrow 5 } f(x) = 0 \)

 

Q28.

Suppose \(f(x) = \left\{\begin{matrix} a + bx, & if \; x < 1 \\ 4, & if \; x = 0, \\ b – ax, & if \; x > 1 \end{matrix}\right.\)

And \(\lim_{x \rightarrow 1} f(x) = f(1)\) what can be the values of b and a?

Soln:

Given function

\(f(x) = \left\{\begin{matrix} a + bx, & if \; x < 1 \\ 4, & if \; x = 0, \\ b – ax, & if \; x > 1 \end{matrix}\right.\) \(\lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1}(a + bx) = a + b\) \(\lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1}(b – ax) = b – a \)

f(1) = 4

Given that \(\lim_{x \rightarrow 1} f(x) = f(1)\).

\(∴ \lim_{x \rightarrow 1^{-}}f(x) = \lim_{x \rightarrow 1^{+}}f(x) = \lim_{x \rightarrow 1} f(x) = f(1)\)

=> a+ b = 4 and b – a = 4

Solving both the equations, we get a = 0 and b = 4.

Hence, b and a are 4 and 0

 

Q29.

A function \(f(x) = (x – a_{1})(x – a_{2}) . . . . (x – a_{n})\) is define by fixed real numbers a1, a2, . . . . an.

Find \(\lim_{x \rightarrow a_{1}}f(x)\). For some \(a \neq a_{1}, a_{2}, . . . . , a_{n} \; compute \; \lim_{x \rightarrow a_{n}} f(x)\)

Soln:

Given function = \(f(x) = (x – a_{1})(x – a_{2}) . . . (x – a_{n})\)

\(\lim_{x \rightarrow a_{1}}f(x) = \lim_{x \rightarrow a_{1}}[(x – a_{1})(x – a_{2}) . . . (x – a_{n})]\)

 

\(= [\lim_{x \rightarrow a_{1}}(x – a_{1})][\lim_{x \rightarrow a_{1}}(x – a_{2})] . . . [\lim_{x \rightarrow a_{1}}(x – a_{n})]\)

 

\(= \left ( a_{1} – a_{1} \right ) \left ( a_{1} – a_{2} \right ) . . . . (a_{1} – a_{n}) = 0\)

 

\(∴ \lim_{x \rightarrow a_{1}]}f(x) = 0\)

 

Now, \(\lim_{x \rightarrow a_{1}]}f(x) = \lim_{x \rightarrow a}[\left (x – a_{1} \right )\left ( x – a_{2} \right ) . . . \left ( x – a_{n} \right )]\)

 

= \(= \lim_{x \rightarrow a}[ x – a_{1} ]  [ x – a_{2}] . . . [x – a_{n}]\)

 

\(( a_{1} – a_{1}) ( a_{1} – a_{2} ) . . . . (a_{1} – a_{n})\)
\( ∴ \lim_{x \rightarrow a} f(x) =( a_{1} – a_{1})( a_{1} – a_{2}) . . . . (a_{1} – a_{n}) \)

 

Q30.

If \(f(x) = \left\{\begin{matrix} \left | x \right | + 1, & x \leq 0\\ 0, & x = 0 \\ \left | x \right | – 1, & x > 0 \end{matrix}\right.\)

\(\lim_{x \rightarrow a} f(x)\) will exist for what values?

Soln:

Given function:

\(f(x) = \left\{\begin{matrix} \left | x \right | + 1, & x \leq 0\\ 0, & x = 0 \\ \left | x \right | – 1, & x > 0 \end{matrix}\right.\)

When a = 0

\(\lim_{x \rightarrow 0^{-} f(x)} = \lim_{x \rightarrow 0^{-}}\left ( \left | x \right | + 1 \right )\)

\(= \lim_{x \rightarrow 0}\left ( – x + 1 \right )\)   [if x < 0, |x| = -x]

= -0 + 1

= 1

\(\lim_{x \rightarrow 0^{+} f(x)} = \lim_{x \rightarrow 0^{+}}\left ( \left | x \right | – 1 \right )\)

\(= \lim_{x \rightarrow 0}\left ( x – 1 \right )\)   [if x > 0, |x| = x]

= 0 – 1

= -1

We have, \(\lim_{x \rightarrow 0^{-}} f(x) \neq \lim_{x \rightarrow 0^{+}} f(x)\)

\(∴ \lim_{x \rightarrow 0} f(x)\) doesn’t exist.

When a < 0

\(\lim_{x \rightarrow a^{-}}f(x) = \lim_{x \rightarrow a^{-}}(\left | x \right | + 1)\) \(= \lim_{x \rightarrow a}(- x + 1) \; \; \; \; [ x < a < 0 \Rightarrow \left | x \right | = – x ]\)

= -a + 1

\(\lim_{x \rightarrow a^{+}}f(x) = \lim_{x \rightarrow a^{+}}(\left | x \right | + 1)\) \(= \lim_{x \rightarrow a}(- x + 1) \; \; \; \; [ x < a < 0 \Rightarrow \left | x \right | = – x ]\)

= – a + 1

\(∴ \lim_{x \rightarrow a^{-} } = \lim_{x \rightarrow a^{+} } = – a + 1\)

Hence, at x = a limit of f(x) exist, where a > 0

Thus, \(\lim_{x \rightarrow a} f(x) \; exist \; for \; all \; a \neq 0\)

 

Q31.

Evaluate \(\lim_{x \rightarrow 1} f(x)\)

If f(x) satisfies, \(\lim_{x \rightarrow 1} \frac{f(x) – 2}{x^{2} – 1} = \pi\)

Soln:

\(\lim_{x \rightarrow 1} \frac{f(x) – 2}{x^{2} – 1} = \pi\) \( \Rightarrow \frac{\lim_{x \rightarrow 1} \left ( f(x) – 2 \right )}{ \lim_{x \rightarrow 1}  \left ( x^{2} – 1 \right )} = \pi\) \(\Rightarrow \lim_{x \rightarrow 1} (f(x) – 2) = \pi \lim_{x \rightarrow 1} (x^{2} – 1)\) \(\Rightarrow \lim_{x \rightarrow 1} (f(x) – 2) = \pi (1^{2} – 1)\) \(\Rightarrow \lim_{x \rightarrow 1} (f(x) – 2) = 0 \) \(\Rightarrow \lim_{x \rightarrow 1} f(x) – \lim_{x \rightarrow 1} 2 = 0 \) \(\Rightarrow \lim_{x \rightarrow 1} f(x) – 2 = 0 \) \(∴ \lim_{x \rightarrow 1} f(x) = 2\)

 

Q32.

If \(f(x) = \left\{\begin{matrix} ax^{2} + b, & x < 0 \\ bx + a, & 0 \leq x \leq 1 \\ bx^{3} + a & x > 1 \end{matrix}\right.\)

For what values of a and b does \(\lim_{x \rightarrow 0} f(x) \; and \; \lim_{x \rightarrow 1} f(x)\) exist?

 

Soln:

Given function

\(f(x) = \left\{\begin{matrix} ax^{2} + b, & x < 0 \\ bx + a, & 0 \leq x \leq 1 \\ bx^{3} + a & x > 1 \end{matrix}\right.\) \(\lim_{x \rightarrow 0^{-} f(x)} = \lim_{x \rightarrow 0}(ax^{2} + b)\)

= a(0)2 + b

= b

\(\lim_{x \rightarrow 0^{+}} f(x) = \lim_{x \rightarrow 0}(bx + a)\)

= b(1) + a

= a + b

\(∴ \lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1^{+}} f(x) = \lim_{x \rightarrow 1} f(x)\)

Thus, for integral values of n and m \(\lim_{x \rightarrow 1}f(x)\) exist.

 

 

Exercise-13.2

Q1. Find derivative for x2 – 2 at x = 10

 

Soln:

Let f(x) = x2 – 2

Accordingly,

\(f'(10) = \lim_{h \rightarrow 0} \frac {f(10 + h) – f(10)}{h}\)

 

\(= \lim_{h \rightarrow 0} \frac {[(10 + h)^{2} – 2] – (10^{2} – 2)}{h}\)

 

\(= \lim_{h \rightarrow 0} \frac {10^{2} + 2. 10. h + h^{2} – 2 – 10^{2} + 2}{h}\)

 

\(= \lim_{h \rightarrow 0} \frac {20h + h^{2}}{h} \)

 

\(= \lim_{h \rightarrow 0} (20 + h) = (20 + 0) = 0 \)

 

Hence, derivative for x2 – 2 at x = 10 is 20

 

 

Q2. Find derivative

99x at x = 100

 

Soln:

Let f(x) = 99x,

Accordingly,

\(f'(100) = \lim_{h \rightarrow 0} \frac{f(100 + h) – f(100)}{h}\)

 

\( = \lim_{h \rightarrow 0} \frac{99(100 + h) – 99(100)}{h}\)

 

\( = \lim_{h \rightarrow 0} \frac{99 \times 100 + 99h – 99 \times 100}{h}\)

 

\( = \lim_{h \rightarrow 0} \frac{ 99h }{h}\)

 

\( = \lim_{h \rightarrow 0} (99h) = 99 \)

 

Hence, derivative for 99x at x = 100 is 99

 

 

Q3. Find derivative

x at x = 1

 

Soln:

Let f(x) = x

Accordingly,

\(f'(1) = \lim_{h \rightarrow 0} \frac{f(1 + h) – f(1)}{h}\)

 

\(= \lim_{h \rightarrow 0} \frac{(1 + h) – 1}{h}\)

 

\(= \lim_{h \rightarrow 0} \frac{h}{h}\)

 

\(= \lim_{h \rightarrow 0} (1)\)

 

= 1

Hence, derivative for x at x = 1 is 1

 

 

Q4. Using first principle find derivative

(i) x3 – 27

 

(ii) (x – 1) (x – 2)

 

(iii) \(\frac{1}{x^{2}}\)

 

(iv) \(\frac{x + 1}{x – 1}\)

Soln:

(i) Let f(x) = x3 – 27

From first principle,

\(f'(x) = \lim_{h \rightarrow 0}\frac{f(x + h) – f(x)}{h}\)

 

\( = \lim_{h \rightarrow 0}\frac{ [(x + h)^{3} – 27] – (x^{3} – 27)} {h} \)

 

\( = \lim_{h \rightarrow 0}\frac{ x^{3} + h^{3} + 3x^{2}h + 3xh^{2} – x^{3}}{h} \)

 

\( = \lim_{h \rightarrow 0}\frac{ h^{3} + 3x^{2}h + 3xh^{2}} {h} \)

 

\( = \lim_{h \rightarrow 0} \left ( h^{2} + 3x^{2} + 3xh \right ) \)

 

= 0 + 3x2 + 0 = 3x2

 

(ii) Let f(x) = (x – 1)(x – 2)

From first principle,

\(f'(x) = \lim_{h \rightarrow 0}\frac{f(x + h) – f(x)}{h}\)

 

\(= \lim_{h \rightarrow 0}\frac{(x + h – 1) (x + h – 2) – (x – 1) (x – 2)} {h}\)

 

\(= \lim_{h \rightarrow 0}\frac{(x^{2} + hx – 2x + hx + h^{2} – 2h – x – h + 2 ) – (x^{2} – 2x – x + 2)} {h}\)

 

\(= \lim_{h \rightarrow 0}\frac{(hx + hx + h^{2} – 2h -h )} {h}\)

 

\(= \lim_{h \rightarrow 0}\frac{( 2hx + h^{2} – 3h )} {h}\)

 

\(= \lim_{h \rightarrow 0} ( 2x + h – 3 )\)

 

\( ( 2x + h – 3 )\)

 

= 2x – 3

 

(iii) Let \(f(x) = \frac{1}{x^{2}}\)

From first principle

\(f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) – f(x)}{h}\)

 

\( = \lim_{h \rightarrow 0} \frac{\frac{1}{(x + h)^{2}} – \frac{1}{x^{2}}}{h}\)

 

\( = \lim_{h \rightarrow 0}  \frac{1}{h} \left [\frac{ x^{2} – (x + h )^{2}}{x^{2} (x + h )^{2}}\right ]\)

 

\( = \lim_{h \rightarrow 0}  \frac{1}{h} \left [\frac{ x^{2} – x^{2} – h^{2} – 2hx}{x^{2} (x + h )^{2}}\right ]\)

 

\( = \lim_{h \rightarrow 0} \frac{1}{h} \left [\frac{ – h^{2} – 2hx}{x^{2} (x + h )^{2}}\right ]\)

 

\( = \lim_{h \rightarrow 0} \left [\frac{ – h – 2hx}{x^{2} (x + h )^{2}}\right ]\)

 

\(\frac{0 – 2x}{x^{2}(x + 0)^{2}} = \frac{-2}{x^{3}}\)

 

(iv) Let \(f(x) = \frac{x +1}{x – 1}\)

From first principle,

\( = \lim_{h \rightarrow 0} \frac{\left ( \frac{x + h + 1}{x + h – 1} – \frac{x + 1}{x – 1} \right )}{h}\)

 

\( = \lim_{h \rightarrow 0} \frac {1}{h} \left [ \frac{\left ( x – 1 \right ) \left ( x+ h + 1 \right ) – \left ( x + 1 \right ) \left ( x + h – 1 \right )}{\left ( x – 1 \right ) \left ( x+ h – 1 \right )} \right ]\)

 

\( = \lim_{h \rightarrow 0} \frac {1}{h} \left [ \frac{\left ( x^{2} + hx + x – x – h – 1 \right ) – \left ( x^{2} + hx – x + x + h – 1 \right )}{\left ( x- 1 \right )\left ( x + h + 1 \right )} \right ]\)

 

\( = \lim_{h \rightarrow 0} \frac {1}{h} \left [ \frac{-2h}{\left ( x – 1 \right ) \left ( x+ h – 1 \right )} \right ]\)

 

\( = \lim_{h \rightarrow 0} \left [ \frac{-2}{\left ( x – 1 \right ) \left ( x+ h – 1 \right )} \right ]\)

 

\(= \frac{-2}{\left ( x – 1 \right )\left ( x – 1 \right )} = \frac{-2}{\left ( x – 1 \right )^{2}}\)

 

 

Q5. Prove \(f'(1) = 100 f'(0)\)

For the function \(f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + … + \frac{x^{2}}{2} + x + 1\)

 

Soln:

Given function

\(f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + … + \frac{x^{2}}{2} + x + 1\)

 

\(\frac{d}{dx} f(x) = \frac{d}{dx}\left [ \frac{x^{100}}{100} + \frac{x^{99}}{99} + … + \frac{x^{2}}{2} + x + 1 \right ]\)

 

\(\frac{d}{dx} f(x) = \frac{d}{dx} \left ( \frac{x^{100}}{100} \right ) + \frac{d}{dx}\left ( \frac{x^{99}}{99} \right ) + … + \frac{d}{dx}\left ( \frac{x^{2}}{2} \right ) + \frac{d}{dx}\left ( x \right )+ \frac{d}{dx}\left ( 1 \right )\)

 

Using theorem \(\frac{d}{dx} \left ( x^{n} \right ) = n x^{n – 1}, \) we get

 

\(\frac{d}{dx} f(x) = \frac{100x^{99}}{100} + \frac{99x^{98}}{99} + . . . . + \frac{2x}{2} + 1 + 0\)

 

\(= x^{99} + x^{98} + …. + x + 1\)

 

∴ \(\frac{d}{dx} f(x) = x^{99} + x^{98} + …. + x + 1\)

 

At x = 0

 

f'(0) = 1

 

At x = 1,

 

f'(1) = 199 + 198 + . . . . + 1 + 1 = [1 + 1 +…… + 1 +1 ]100 terms = 1 x 100 = 100

 

Hence,  f’(1) = 100 x f1 (0)

 

 

Q6. For a real number ‘a’ find the derivative of xn + axn – 1 + a2xn – 2 + . . . .+ an – 1 x + an

Soln:

Let \(f(x) = x ^{n} +  ax ^{n – 1} + a^{2} x ^{n – 2} + . . . + a^{n – 1}x + a^{n}\)

 

\(f’(x) = \frac{d}{dx} \left ( x ^{n} + ax ^{n – 1} + a^{2} x ^{n – 2} + . . . + a^{n – 1}x + a^{n} \right ) \)

 

\(= \frac{d}{dx} \left ( x ^{n} \right ) + a\frac{d}{dx} \left ( x ^{n – 1} \right ) + a^{2} \frac{d}{dx} \left ( x ^{n – 2} \right ) + . . . + a^{n – 1}\frac{d}{dx} \left ( x \right ) + a^{n} \frac{d}{dx} \left ( 1 \right )\)

 

Using \(\frac{d}{dx}x^{n} = nx^{n -1}\), we have

 

\(f'(x) = nx^{n – 1} + a(n – 1)x^{n – 2} + a^{2} (n – 2)x^{n -3} + … + a^{ n – 1} + a^{n}(0)\)

 

\(= nx^{n – 1} + a(n – 1)x^{n – 2} + a^{2} (n – 2)x^{n -3} + … + a^{ n – 1}\)

 

 

Q7. Find the derivative for

(i) (x – m)(x – n)

 

(ii) (ax2 + b)2

 

(iii) \(\frac{x – a}{x – b}\)

Soln:

(i) Let f(x) = (x – m)(x – n)

 

\(\Rightarrow f(x) = x^{2} – (m + n) x + mn\)

 

\(f'(x) = \frac{d}{dx}(x^{2} – (m + n)x + mn)\)

 

\(= \frac{d}{dx}(x^{2}) – (m + n) \frac{d}{dx}(x) + \frac{d}{dx}(mn)\)

 

Using \(= \frac{d}{dx}(x^{n}) = nx^{n – 1}\), we get

 

\(f'(x) = 2x -(m + n) + 0 = 2x – m – n\)

 

(ii) Let f(x) = (ax2 + b)2

 

\(\Rightarrow f(x) = a^{2}x^{2} + 2abx + b^{2}\)

 

\(f'(x) = \frac{d}{dx} (a^{2}x^{4} + 2abx^{2} + b^{2}) = a^{2}\frac{d}{dx}(x^{4}) + 2ab\frac{d}{dx}(x^{2}) + \frac{d}{dx}(b^{2})\)

 

Using \(\frac{d}{dx}x^{n} = nx^{n – 1}\), we have

 

\(f’\left ( x \right ) = a ^{2}(4x^{3}) + 2ab (2x) + b^{2}(0)\)

 

= 4a2x3 + 4abx

 

= 4ax(ax2 + b)

 

(iii) Let \(f(x) = \frac{(x – a)}{(x – b)}\)

 

\(f'(x) = \frac{d}{dx} \left (\frac{x – a}{x – b} \right )\)

 

Quotient rule,

\(f'(x) = \frac {(x – b)\frac{d}{dx}(x – a) – (x – a)\frac{d}{dx}(x- b)}{(x – b)^{2}}\)

 

\(= \frac {(x – b)(1) – (x – a)(1)}{(x – b)^{2}}\)

 

\(= \frac {x – b – x + a}{(x – b)^{2}}\)

 

\(= \frac { a – b }{(x – b)^{2}}\)

 

 

Q8. If a is constant find derivative of \(\frac {x^{n} – a^{n}}{x – a}\)

Soln:

Let \(\frac {x^{n} – a^{n}}{x – a}\)

 

\(\Rightarrow f'(x) = \frac{d}{dx}\left ( \frac {x^{n} – a^{n}}{x – a} \right )\)

 

By Question rule

\(\Rightarrow f'(x) = \frac{(x – a) \frac{d}{dx}(x^{n} – a^{n}) – (x^{n} – a^{n})\frac{d}{dx}(x – a)}{(x – a)^{2}}\)

 

\(= \frac{(x -a)(nx^{n – 1} – 0) – (x^{n} – a^{n})}{(x – a)^{2}}\)

 

\(= \frac{nx^{n} – anx^{n – 1} – x^{n} + a^{n}}{(x – a)^{2}}\)

 

 

Q9.

(i) \( 2x – \frac {3}{4} \)

 

(ii) (5x3 + 3x – 1) (x – 1)

 

(iii) x-3 (5 + 3x)

 

(iv) x5 (3 – 6x-9)

 

(v) x-4 (3 – 4x-5)

 

(vi) \(= \frac{2}{x + 1} – \frac{x^{2}}{3x – 1}\)

 

Soln.

(i) Let \( f(x) = 2x – \frac {3}{4} \)

 

\(f'(x) = \frac{d}{dx}\left ( 2x – \frac{3}{4} \right )\)

 

\(= 2\frac{d}{dx}\left ( x \right ) – \frac {d}{dx} \left ( \frac{3}{4} \right )\)

 

= 2 – 0

 

= 2

 

(ii) Let f(x) = (5x3 + 3x – 1) (x – 1)

By Leibnitz product rule,

\(f'(x) = (5x^{3} + 3x – 1) \frac{d}{dx}(x – 1) + (x – 1) \frac{d}{dx}(5x^{3} + 3x – 1 )\)

 

= (5x3 + 3x – 1) (1) + (x – 1)(5.3x2 + 3 – 0)

 

= (5x3 + 3x – 1) + (x – 1)(15x2 + 3)

 

= 5x3 + 3x – 1 + 15x3 + 3x – 15x2 – 3

 

= 20x3 – 15x2 + 6x – 4

 

(iii) Let f(x) = x-3 (5 + 3x)

 

\(f'(x) = x^{-3}\frac{d}{dx}(5 + 3x) + (5 + 3x)\frac{d}{dx}(x^{-3})\)

 

\( = x^{-3}(0 + 3) + (5 + 3x) (-3x^{-3 – 1} )\)

 

\( = x^{-3}( 3) + (5 + 3x) (-3x^{-4} )\)

 

= 3x-3 – 15x-4 – 9x-3

 

= -6x-3 – 15x-4

 

\(= -3 x ^{-3} \left ( 2 + \frac{5}{x} \right )\)

 

\(= \frac{-3 x ^{-3} }{x}\left ( 2x + 5 \right )\)

 

\(= \frac{-3 }{ x ^{4} }\left ( 2x + 5 \right )\)

 

(iv) let f(x) = x5 (3 – 6x-9)

 

From Leibnitz product rule,

\(f'(x) = x^{5} \frac{d}{dx}(3 – 6x^{-9}) + (3 – 6x^{-9}) \frac{d}{dx}(x^{5})\)

 

\(= x^{5} \left \{ 0 – 6 (-9)x^{-9 – 1} \right \} + (3 – 6x^{-9})(5x^{4})\)

 

\(= x^{5} (54x ^{-10}) + 15x^{4} – 30x^{-5}\)

 

= 54x-5 + 15x4 – 30x-5

 

= 24x-5 + 15x4

 

= \(= 15x^{4} + \frac{24}{x^{5}}\)

 

(v) Let f(x) = x-4 (3 – 4x-5)

 

From Leibnitz product rule,

\(f'(x) = x^{-4}\frac{d}{dx}\left ( 3 – 4x^{-5} \right ) + (3 – 4 x^{-5})\frac{d}{dx}(x^{-4})\)

 

\(= x^{-4}\left \{ 0 – 4 (-5)x^{-5 – 1} + (3 – 4x^{-5})(-4)x^{-4 -1} \right \}\)

 

= x-4 (20x-6) + (3 – 4x-5)(-4x-5)

 

= 36x-10 – 12x-5

 

\(= -\frac{12}{x^{5}} + \frac{36}{x^{10}}\)

 

(iv) Let \( f(x) = \frac{2}{x + 1} – \frac{x^{2}}{3x – 1}\)

\(f'(x) = \frac{d}{dx}\left ( \frac{2}{x + 1} \right ) – \frac{d}{dx} \left ( \frac{x^{2}}{3x – 1} \right )\)

 

From quotient rule,

\(f'(x) = \left [ \frac{\left ( x + 1 \right ) \frac{d}{dx}(2) – 2 \frac{d}{dx} (x + 1) }{(x + 1)^{2}} \right ] – \left [ \frac{(3x – 1) \frac{d}{dx} (x^{2}) – x^{2} \frac{d}{dx}\left ( 3x – 1 \right )}{(3x – 1)^{2}} \right ]\)

 

\(= \left [ \frac{(x + 1)(0) – 2(1)}{(x + 1)^{2}} \right ] – \left [ \frac{(3x – 1)(2x) – (x^{2})(3)}{(3x – 1)^{2}} \right ]\)

 

\(= \frac{-2}{(x + 1)^{2}} – \left [ \frac{6x^{2} – 2x – 3x^{2}}{(3x – 1)^{2}} \right ]\)

 

\(= \frac{-2}{(x + 1)^{2}} – \left [ \frac{3x^{2} – 2x} {(3x – 1)^{2}} \right ]\)

 

\(= \frac{-2}{(x + 1)^{2}} – \frac{x(3x – 2)} {(3x – 1)^{2}}\)

 

 

Q10. Using first principle find derivative of \(\cos x\)

Soln:

Let \(f(x) = \cos x\).

 

According to first principle

\(f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) – f(x)}{h}\)

 

\(= \lim_{h \rightarrow 0} \frac{\cos (x + h) – \cos x}{h}\)

 

\(= \lim_{h \rightarrow 0} \left [\frac{\cos x \cos h – \sin x \sin h – \cos x}{h} \right ]\)

 

\(= \lim_{h \rightarrow 0} \left [\frac{ – \cos x (1 – \cos h) – \sin x \sin h }{h} \right ]\)

 

\(= \lim_{h \rightarrow 0} \left [\frac{ – \cos x (1 – \cos h)}{h} – \frac{\sin x \sin h }{h} \right ]\)

 

\(=- \cos x \left ( \lim_{h \rightarrow 0} \frac{1 – \cosh}{h} \right ) – \sin x\lim_{h \rightarrow 0}\left ( \frac{sin h}{h} \right )\)

 

= \(= – \cos x (0) – \sin x (1)\)   \(\left [ \lim_{h \rightarrow 0} \frac{1 – cos h}{h} = 0 \; and \; \lim_{h \rightarrow 0}\frac{\sin h }{h} = 1 \right ]\)

 

= – sin x

 

f’(x) = – sin x

 

 

Q11.

(i) sin x cos x

 

(ii) sec x

 

(iii) 5 sec x + 4 cos x

 

(iv) cosec x

 

(v) 3cot x + 5cosec x

 

(vi) 5sin x – 6cos x + 7

 

(vii) 2tan x – 7sec x

Soln:

(i) Let f’(x) = sin x cos x

 

According to first principle,

\(f'(x) = \lim_{h \rightarrow 0}\frac{f(x + h) -f(x)}{h}\)

 

\(= \lim_{h \rightarrow 0}\frac{\sin (x + h) \cos(x + h) – \sin x \cos x}{h}\)

 

\(= \lim_{h \rightarrow 0}\frac{1}{2h} \left [ 2\sin (x + h) \cos (x + h) – 2 \sin x cos x \right ]\)

 

\(= \lim_{h \rightarrow 0}\frac{1}{2h} \left [ \sin 2(x + h) – \sin 2x \right ]\)

 

\(= \lim_{h \rightarrow 0}\frac{1}{2h} \left [ 2 \cos \frac{2x + 2h + 2x}{2} . \sin \frac{2x + 2h – 2x}{2} \right ]\)

 

\(= \lim_{h \rightarrow 0}\frac{1}{h} \left [ \cos \frac{ 4x + 2h }{2}  \sin \frac{ 2h }{2} \right ]\)

 

\(\lim_{h \rightarrow 0}\frac{1}{h} \left [ \cos (2x + h ) \sin h \right ]\)

 

\(\lim_{h \rightarrow 0} \cos (2x + h ) \lim_{h \rightarrow 0} \frac{\sin h}{h}\)

 

= cos(2x + 0). 1

 

= cos 2x

 

(ii) Let f(x) = sec x

According to first principle,

\(f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) – f(x)}{h}\)

 

\(= \lim_{h \rightarrow 0} \frac{\sec (x + h) – \sec x}{h}\)

 

\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{1}{\cos (x + h)} – \frac{1}{\cos x} \right ]\)

 

\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\cos x – \cos(x + h)}{\cos x \cos(x + h)} \right ]\)

 

\(= \frac{1}{\cos x} \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{-2 \sin \left ( \frac{x + x + h}{2} \right ) \sin \left ( \frac{x – x – h}{2} \right )}{\cos (x + h)} \right ]\)

 

\(= \frac{1}{\cos x} \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{-2 \sin \left ( \frac{2x + h}{2} \right ) \sin \left ( \frac{- h}{2} \right )}{\cos (x + h)} \right ]\)

 

\(= \frac{1}{\cos x} \lim_{h \rightarrow 0} \frac{\left [ \sin \left ( \frac{2x + h}{2} \right ) \frac{\sin \left ( \frac{h}{2}\right )}{\left (\frac{h}{2}\right )} \right ]}{\cos (x + h)}\)

 

\(= \frac{1}{\cos x} \lim_{h \rightarrow 0} \frac{\sin \left ( \frac{h}{2} \right )}{\left ( \frac{h}{2} \right )}. \lim_{h \rightarrow 0}\frac{\sin\left ( \frac{2x + h}{2} \right ) }{\cos (x + h)}\)

 

\(= \frac{1}{\cos x} . 1 . \frac{\sin x}{ \cos x}\)

 

\(= \sec x \tan x\)

 

(iii)

Let f(x) = 5 sec x + 4 cos x

 

According to first principle,

\(f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) – f(x)}{h}\)

 

\(\lim_{h \rightarrow 0}\frac{5\sec (x + h) + 4 \cos(x + h) – \left [ 5 \ sec x + 4 \ cos x \right ]}{h}\)

 

\(= 5\lim_{h \rightarrow 0}\frac{[\sec(x + h) – \sec x]}{h} + 4 \lim_{h \rightarrow 0}\frac{[\cos(x + h) – \cos x]}{h}\)

 

\(= 5\lim_{h \rightarrow 0}\frac{1}{h}\left [ \frac{1}{\cos (x + h)} – \frac{1}{\cos x} \right ] + 4 \lim_{h \rightarrow 0}\frac{1}{h} \left [ \cos (x + h) – \cos x \right ]\)

 

\(= 5\lim_{h \rightarrow 0}\frac{1}{h}\left [ \frac{\cos x – \cos (x + h)}{\cos x \cos (x + h)} \right ] + 4 \lim_{h \rightarrow 0}\frac{1}{h}\left [ \cos x \cos h – \sin x \sin h – \cos x \right ]\)

 

\(\\= \frac{5}{\cos x} \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{-2\sin \left ( \frac{x + x + h}{2} \right )\sin \left ( \frac{x – x – h}{2} \right )}{cos(x + h)} \right ] + 4 \lim_{h\rightarrow 0}\frac{1}{h} [-\cos x (1 – \cos h) – \sin x sin h]\)

 

\(\\= \frac{5}{\cos x} \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{-2\sin \left ( \frac{ 2x + h}{2} \right )\sin \left ( \frac{ – h}{2} \right )}{cos(x + h)} \right ] + 4\left [ – \cos x \lim_{h \rightarrow 0} \frac{(1 – \cos h)}{h} – \sin x \lim_{h \rightarrow 0} \frac{\sin h}{h} \right ] \)

 

\(= \frac{5}{\cos x}. \lim_{h \rightarrow 0} \left [ \frac{\sin\left ( \frac{2x + h}{2} \right ). \frac{\sin \left ( \frac{h}{2} \right )}{\frac{h}{2}}}{\cos (x + h)}\right ] + 4 [(-\cos x).(0) – (\sin x). 1]\)

 

\(= \frac{5}{\cos x}. \left [ \lim_{h \rightarrow 0} \frac{\sin\left ( \frac{2x + h}{2} \right )}{\cos (x + h)}. \lim_{h \rightarrow 0} \frac{\sin \left ( \frac{h}{2} \right )}{\frac{h}{2}}\right ] + 4 \sin x\)

 

\(= \frac{5}{\cos x}. \frac{\sin x}{\cos x}. 1 – 4 \sin x\)

 

\(= 5 \sec x \tan x – 4 \sin x\)

 

(iv)

Let f(x) = cosec x

 

According to first principle

\(f'(x) = \lim_{h \rightarrow 0}\frac{f(x + h) – f(x)}{h}\)

 

\(f'(x) = \lim_{h \rightarrow 0} \frac{1}{h} \left [ \csc (x + h) – \csc x \right ] \)

 

\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{1}{\sin (x + h)} – \frac{1}{\sin x} \right ]\)

 

\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\sin x – \sin(x + h)}{ \sin (x + h) \sin x} \right ]\)

 

\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{2 \cos \left ( \frac{x + x + h}{2} \right ). \sin \left ( \frac{x – x – h}{2} \right ) }{\sin (x + h) . \sin x} \right ]\)

 

\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{2 \cos \left ( \frac{ 2x + h}{2} \right ). \sin \left ( \frac{- h}{2} \right ) }{\sin (x + h) . \sin x} \right ]\)

 

\(= \lim_{h \rightarrow 0} \frac{ – \cos \left ( \frac{ 2x + h}{2} \right ). \frac{\sin \left ( \frac{ h}{2} \right )}{\left ( \frac{ h}{2} \right )} }{\sin (x + h) . \sin x}\)

 

\(= \lim_{h \rightarrow 0} \left ( \frac{ – \cos \left ( \frac{ 2x + h}{2} \right ). }{\sin (x + h) . \sin x} \right ) \lim_{\frac{h}{2} \rightarrow 0} \frac{\sin \left ( \frac{ h}{2} \right )}{\left ( \frac{ h}{2} \right )}\)

 

\(= \left ( \frac{-\cos x}{\sin x \sin x} \right ).1\)

 

= – cosec x cot x

 

(v)

Let f(x) = 3cot x + 5cosec x

 

According to first principle,

\(f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) – f(x)}{h}\)

 

\(= \lim_{h \rightarrow 0} \frac{3 \cot (x + h) + 5 \csc (x + h) – 3 \cot x – 5 \csc x}{h}\)

 

\(= 3 \lim_{h \rightarrow 0} \frac{1}{h} [\cot (x + h) – \cot x] + 5 \lim_{h \rightarrow 0} \frac{1}{h} [\cot (x + h) – \csc x]\) ….. (1)

 

Now,

\( \lim_{h \rightarrow 0} \frac{1}{h} [\cot (x + h) – \cot x] \)

 

= \( \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\cos (x + h)}{\sin (x + h)} – \frac{\cos x}{\sin x} \right ] \)

 

= \(\lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\cos(x + h)\sin x – \cos x \sin (x + h)}{\sin x sin(x + h)} \right ]\)

 

= \(\lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\sin (x – x – h)}{\sin x sin(x + h)} \right ]\)

 

= \(\lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\sin (- h)}{\sin x sin(x + h)} \right ]\)

 

= \(– \left ( \lim_{h \rightarrow 0} \frac{\sin h}{h} \right ). \left ( \lim_{h \rightarrow 0} \frac{1}{\sin x. \sin (x + h)} \right )\)

 

= \(– 1 . \frac{1}{\sin x . \sin (x + 0)} = \frac{-1}{\sin^{2}x} = -\csc ^{2}x\) ……. (2)

 

\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \csc (x + h) – \csc x \right ]\)

 

\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{1}{\sin (x + h)} – \frac{1}{\sin x} \right ]\)

 

\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\sin x – \sin (x + h) }{\sin x \sin (x + h)} \right ]\)

 

\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{2 \cos \left ( \frac{x + x + h}{2} \right ) . \sin \left ( \frac{x – x – h}{2} \right )}{\sin (x + h) \sin x} \right ]\)

 

\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{2 \cos \left ( \frac{ 2x + h}{2} \right ) . \sin \left ( \frac{- h}{2} \right )}{\sin (x + h) \sin x} \right ]\)

 

\(= \lim_{h \rightarrow 0} \frac{ -\cos \left ( \frac{ 2x + h}{2} \right ) . \frac{\sin \left ( \frac{ h}{2} \right )}{\left ( \frac{ h}{2} \right )}}{\sin (x + h) \sin x}\)

 

\(= \lim_{h \rightarrow 0} \left ( \frac{ -\cos \left ( \frac{ 2x + h}{2} \right )} {\sin (x + h) \sin x} \right ) \lim_{\frac{h}{2} \rightarrow 0} \frac{\sin \left ( \frac{ h}{2} \right )}{\left ( \frac{ h}{2} \right )}\)

 

\(= \left ( \frac{- \cos x }{\sin x \sin x} \right ) . 1\)

 

\(= – \csc x \cot x\) ………. (3)

 

From eqn (1), (2), and (3), we obtain

\(f'(x) = – 3 \csc^{2} x – 5 \csc x \cot x\)

 

(vi)

Let f(x) = 5sin x – 6cos x + 7

 

According to first principle,

\(f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) – f(x)}{h}\)

 

\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ 5 \sin(x + h) – 6 \cos (x + h) + 7 – 5 \sin x + 6 \cos x – 7 \right ]\)

 

\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ 5 \left \{ \sin(x + h) – \sin x \right \} – 6 \left \{ \cos (x + h) – \cos x \right \} \right ]\)

 

\(= 5 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \sin(x + h) – \sin x \right ] – 6 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \cos (x + h) – \cos x \right ]\)

 

\(= 5 \lim_{h \rightarrow 0} \frac{1}{h} \left [ 2 \cos\left ( \frac{x + x + h}{2} \right )  \sin \frac{x + h – x}{2} \right ] – 6 \lim_{h \rightarrow 0} \frac{\cos x \cos h – \sin x \sin h – \cos x}{h}\)

 

\(= 5 \lim_{h \rightarrow 0} \frac{1}{h} \left [ 2 \cos\left ( \frac{ 2x + h}{2} \right ) \sin \frac{ h }{2} \right ] – 6 \lim_{h \rightarrow 0} \left [  \frac{ – \cos x (1 – \cos h ) – \sin x \sin h }{h} \right ] \)

 

\(= 5 \lim_{h \rightarrow 0} \left ( \cos\left ( \frac{ 2x + h}{2} \right ) \frac{ \sin \frac{ h }{2}}{\frac{ h }{2}} \right ) – 6 \lim_{h \rightarrow 0} \left [ \frac{ – \cos x (1 – \cos h )}{h} – \frac {\sin x \sin h }{h} \right ]\)

 

\(= 5 \left [ \lim_{h \rightarrow 0} \cos\left ( \frac{ 2x + h}{2} \right ) \right ] \left [ \lim_{\frac{h}{2} \rightarrow 0}\frac{ \sin \frac{ h }{2}}{\frac{ h }{2}} \right ] – 6 \left [ (- \cos x ) \left ( \lim_{h \rightarrow 0} \frac{ (1 – \cos h )}{h} \right ) – \sin x \lim_{h \rightarrow 0} \left ( \frac {\sin h }{h} \right ) \right ]\)

 

\(= 5 \cos x. 1 – 6 [(-\cos x). (0) – \sin x.1]\)

 

= 5cos x + 6 sin x

 

(vii)

Let f (x) = 2 tan x – 7 sec x

 

Accordingly to first principle,

\(f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) – f(x)}{h}\)

 

\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ 2 \tan (x + h) – 7 \sec (x + h) – 2 \tan x + 7 \sec x\right ]\)

 

\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ 2 \left \{ \tan (x + h) – \tan x \right \} – 7 \left \{ \sec (x + h) – \sec x \right \} \right ]\)

 

\(= 2 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \tan (x + h) – \tan x \right ] – 7 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \sec (x + h) – \sec x \right ]\)

 

\(= 2 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\sin (x + h)}{\cos (x + h)} – \frac{\sin x}{\cos x} \right ] – 7 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{1}{\cos (x + h)} – \frac{1}{\cos x} \right ]\)

 

\(= 2 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\sin (x + h) \cos x – \sin x \cos (x + h) }{ \cos x \cos (x + h)} \right ] – 7 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\cos x – \cos (x + h) }{\cos x \cos (x + h)} \right ]\)

 

\(= 2 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{ \sin (x + h – x) }{ \cos x \cos (x + h)} \right ] – 7 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{-2 \sin \left ( \frac{x + x + h}{2} \sin \left ( \frac{x – x – h}{2} \right ) \right )}{\cos x \cos (x + h)} \right ]\)

 

\(= 2 \left ( \lim_{h \rightarrow 0} \frac{\sin h}{h} \right ) \left ( \lim_{h \rightarrow 0} \frac{ 1 }{ \cos x \cos (x + h)} \right ) – 7 \left( \lim_{\frac{h}{2} \rightarrow 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}} \right ) \left ( \lim_{h \rightarrow 0} \frac{ \sin \left ( \frac{ 2x + h}{2} \right ) }{\cos x \cos (x + h)} \right )\)

 

\(= 2 . 1 . \frac{1}{\cos x \cos x} – 7 . 1 \left ( \frac{\sin x}{\cos x \cos x} \right )\)

 

\(= 2 \sec^{2} x – 7 \sec x \tan x\)

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