What is a Limit in calculus? A limit can be explained as the value which a function tends to approach as the input value (also known as index) gains some value. It is the converging of different values at a point. Boundedness of a function is shown by limits. In this article, we come across solved examples of limits.

Consider f(x) to be a function. In a function, if x takes a definite value say b, x → b is called limit. Here ‘b’ is a value which is pre-assigned. It is represented as lim_x→bf(x).

The tendency of f(x) at x=a towards the left is called left limit and denoted by lim_x→a^– and towards the right is called right limit denoted by lim_x→a⁺. Limit of a function at a point is the common value of the right and left hand limits, if they coincide.

The graphical representation of limits is as follows:

Graphical Representation of Limits

Algebra of Limits

Suppose lim_x→a a(x) = r and lim_x→a b(x) = t then the following can be defined.

lim_x→a [r(x) + t(x)] = lim_x→a r(x) + lim_x→a t(x).
lim_x→a [r(x) − t(x)] = lim_x→a r(x) − lim_x→a t(x).
lim_x→a [r(x) × t(x)] = [lim_x→a r(x)] × [lim_x→a t(x)].
lim_x→a [r(x) ÷ t(x)] = [lim_x→a r(x)] ÷ [lim_x→a t(x)].
lim_x→a [α.r(x))] = α. lim_x→a r(x).

Some Standard Limits

The following are some of the standard limits.

\begin{array}{l} 1] lim_{x \to 0} [\frac{\sin x}{x}] = 1 \\ 2] lim_{x \to 0} [\frac{\tan x}{x}] = 1 \\ 3] lim_{x \to \infty} {(1 + \frac{1}{x})}^{x} = e \\ 4] lim_{x \to 0} \frac{\ln (1 + x)}{x} = 1 \\ 5] lim_{x \to 0} \frac{\log_{a} (1 + x)}{x} = \frac{1}{\ln a} \\ 6] lim_{x \to 0} \frac{a^{x} - 1}{x} = \ln a \\ 7] lim_{x \to 1} \frac{x^{m} - 1}{x - 1} = m \\ 8] lim_{x \to a} \frac{x^{m} - a^{m}}{x - a} = m a^{m - 1} \\ 9] lim_{x \to 0} \frac{\sin^{- 1} x}{x} = lim_{x \to 0} \frac{\tan^{- 1} x}{x} = 1 \end{array}

Related articles

Limits, Continuity and Differentiability

Limits of Functions

Solved Examples on Limits for Practice

Below are illustrated some of the questions based on limits asked in JEE previous exams.

Example 1: Find lim_x→∞sinx/x.

Solution:

Let x = 1/y or y = 1/x, so that x → ∞ ⇒ y → 0

∴ lim_x→∞(sin x / x) = lim_y→0(y . sin (1 / y))

=lim_y→0y . lim_y→0sin (1 / y)

= 0

Example 2: If f(a) = 2, f′(a) = 1, g(a) = −1; g′(a) = 2, then find

Solved Practice Examples on Limits

Solution:

Consider Practice Examples on Limits

Examples on Limits

Example 3: Evaluate

Solve Limit Problems

Solution:

Limit Problems Solution

Example 4: Find the limit lim_n→∞ [1/n² + 2/n² + … + n/n²]

Solution:

lim_n→∞ [1/n² + 2/n² + … + n/n²]

= lim_n→∞[1+2+3+…. +n]/ n²

= lim_n→∞[(n / 2) * ( n+1)] / n²

= ½ lim_n→∞ ( n+1) / n

= ½ lim_n→∞(1 + 1/n)

= -½

Example 5: Find lim_x→0 sin (π cos²x) / x²

Solution:

Problems on Limits in Calculus

Limits Problems in Calculus

= π (-1).1.(-1) = π

Example 6: let f: R→R be such that f (1) = 3 and f’(1) = 6. Then find the value of

lim _x→0 [f (1+x) / f (1)]^1/x.

Solution:

Let y = [f (1 + x) / f (1)]^1/x

So, log y = 1/x [log f (1 + x) – log f (1)]

So, lim_x→0 log y = lim_x→0[1 / f (1 + x) . f’(1 + x)]

= f’(1) / f(1)

= 6/3

log (lim_x→0 y) = 2

lim_x→0 y = e²

Example 7: If f (x) = [2/x ] − 3, g (x) = x − [3 / x] + 4 and h (x) = −[2 (2x + 1)] / [x²+ x − 12], then what is the value of lim_x→3[f (x) + g (x) + h (x)]?

Solution:

We have f (x) + g (x) + h (x) = [x²− 4x + 17− 4x − 2] / [x²+ x − 12]

= [x²− 8x + 15] / [x²+ x − 12]

= [(x − 3) (x − 5)] / [(x − 3) (x + 4)]

∴lim_x→3[f (x) + g (x) + h (x)] = lim_x→3(x − 3) (x − 5) / (x − 3) (x + 4)

= −2/7

Example 8:

\begin{array}{l} lim_{x \to 0} (\frac{\sqrt{1 - c o s 2 x}}{\sqrt{2} x}) \end{array}

Solution:

\begin{array}{l} If lim_{x \to a -} f (x) \neq lim_{x \to a +} f (x) then the limit does not exist \end{array}

\begin{array}{l} lim_{x \to 0 +} (\frac{\sqrt{1 - \cos (2 x)}}{\sqrt{2} x}) \\ lim_{x \to a} [c \cdot f (x)] = c \cdot lim_{x \to a} f (x) \\ = \frac{1}{\sqrt{2}} \cdot lim_{x \to 0 +} (\frac{\sqrt{1 - \cos (2 x)}}{x}) \\ Apply L^{'} {Hospital}^{'} s Rule \\ = \frac{1}{\sqrt{2}} \cdot lim_{x \to 0 +} (\frac{\frac{\sin (2 x)}{\sqrt{1 - \cos (2 x)}}}{1}) \\ = \frac{1}{\sqrt{2}} \cdot lim_{x \to 0 +} (\frac{\sin (2 x)}{\sqrt{1 - \cos (2 x)}}) \\ Apply L^{'} {Hospital}^{'} s Rule \\ = \frac{1}{\sqrt{2}} \cdot lim_{x \to 0 +} (\frac{\cos (2 x) \cdot 2}{\frac{\sin (2 x)}{\sqrt{1 - \cos (2 x)}}}) \\ Simplify \frac{\cos (2 x) \cdot 2}{\frac{\sin (2 x)}{\sqrt{1 - \cos (2 x)}}} \\ = \frac{1}{\sqrt{2}} \cdot lim_{x \to 0 +} (2 \cot (2 x) \sqrt{1 - \cos (2 x)}) \\ = \frac{1}{\sqrt{2}} \cdot 2 \cdot lim_{x \to 0 +} (\cot (2 x) \sqrt{1 - \cos (2 x)}) \\ Multiply by the conjugate of 1 - \cos (2 x) \\ = \frac{(1 - \cos (2 x)) (1 + \cos (2 x))}{1 + \cos (2 x)} \\ Expand (1 - \cos (2 x)) (1 + \cos (2 x)) \\ = \frac{\sin^{2} (2 x)}{1 + \cos (2 x)} \\ = \frac{1}{\sqrt{2}} \cdot 2 \cdot lim_{x \to 0 +} (\cot (2 x) \sqrt{\frac{\sin^{2} (2 x)}{1 + \cos (2 x)}}) \\ = \frac{1}{\sqrt{2}} \cdot 2 \cdot lim_{x \to 0 +} (\cos (2 x) \sqrt{\frac{1}{\cos (2 x) + 1}}) \\ Plug in the value x = 0 \\ = \frac{1}{\sqrt{2}} \cdot 2 \cos (2 \cdot 0) \sqrt{\frac{1}{\cos (2 \cdot 0) + 1}} \\ Simplify \frac{1}{\sqrt{2}} \cdot 2 \cos (2 \cdot 0) \sqrt{\frac{1}{\cos (2 \cdot 0) + 1}} \\ = 1 \\ Similarly lim_{x \to 0 -} (\frac{\sqrt{1 - \cos (2 x)}}{\sqrt{2} x}) = - 1 \\ = diverges \end{array}

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