NCERT Solutions For Class 11 Maths Chapter 13

For students of class 11 who are looking to improve their knowledge on the chapter, limits and derivatives, we have provided NCERT Solutions for Class 11 Maths Chapter 13. The NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives contains topics such as limits of polynomials and rationals functions, logarithmic, trigonometric as well as exponential functions and more concepts of limits and derivatives. Thus here we have provided downloadable forms of NCERT Solutions for Class 11 Maths Chapter 13 pdf so that students can access the content offline.

NCERT Solutions Class 11 Maths Chapter 13 Exercises

• NCERT Solutions Class 11 Maths Chapter 13 Limits and Derivatives Exercise 13.1
• NCERT Solutions Class 11 Maths Chapter 13 Limits and Derivatives Exercise 13.2

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, . . . . a_n.

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

 

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

 

$( a_{1} – a_{1}) ( a_{1} – a_{2} ) . . . . (a_{1} – a_{n})$
∴limx→af(x)=(a1–a1)(a1–a2)....(a1