# Limits Solved Examples

What is a Limit in calculus? A limit can be explained as the value which a function tends to approach as the input value (which is 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 limits solved examples.

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 limx→bf(x).

The tendency of f(x) at x=a towards the left is called left limit and denote by limx→a and towards the right is called right limit denoted by limx→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:

## Algebra of limits

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

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

## Some Standard Limits

The following are some of the standard limits.

$1] \ \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\sin x}}{x}} \right] = 1\\ 2] \ \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\tan x}}{x}} \right] = 1\\ 3] \ \mathop {\lim }\limits_{x \to \infty } {{\left( {1 + \frac{1}{x}} \right)}^x} = e\\ 4] \ \mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {1 + x} \right)}}{x} = 1\\ 5] \ \mathop {\lim }\limits_{x \to 0} \frac{{{{\log }_a}\left( {1 + x} \right)}}{x} = \frac{1}{{\ln \,a}}\\ 6] \ \mathop {\lim }\limits_{x \to 0} \frac{{{a^x} – 1}}{x} = \ln \,\,a\\ 7] \ \mathop {\lim }\limits_{x \to 1} \frac{{{x^m} – 1}}{{x – 1}} = m\\ 8] \ \mathop {\lim }\limits_{x \to a} \frac{{{x^m} – {a^m}}}{{x – a}} = m\,\,{a^{m – 1}}\\ 9] \ \mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^{ – 1}}x}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{{{\tan }^{ – 1}}x}}{x} = 1\\$

### Related articles

Limits, Continuity and Differentiability

Limits of Functions

## Solved Practice Examples on Limits

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

Example 1: Find limx→∞ sinx/x.

Solution:

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

∴ limx→∞ (sin x / x) = limy→0 (y . sin (1 / y))

=limy→0  y . limy→0 sin (1 / y)

= 0

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

Solution:

Consider

Example 3: Evaluate

Solution:

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

Solution:

limn→∞ [1/n2 + 2/n2 + … + n/n2]

= limn→∞ [1+2+3+…. +n]/ n2

= limn→∞ [(n / 2) * ( n+1)] / n2

= ½ limn→∞ ( n+1) / n

= ½ limn→∞ (1 + 1/n)

= -½

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

Solution:

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

= f’(1) / f(1)

= 6/3

log (limx→0 y) = 2

limx→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 limx→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)]

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

= −2/7

Example 8: $\lim _{x\to 0}\left(\frac{\sqrt{1-cos2x}}{\sqrt{2}x}\right)$

Solution:

$\mathrm{If\:}\lim _{x\to a-}f\left(x\right)\ne \lim _{x\to a+}f\left(x\right)\mathrm{\:then\:the\:limit\:does\:not\:exist}$

$\lim _{x\to \:0+}\left(\frac{\sqrt{1-\cos \left(2x\right)}}{\sqrt{2}x}\right)\\\lim _{x\to a}\left[c\cdot f\left(x\right)\right]=c\cdot \lim _{x\to a}f\left(x\right)\\=\frac{1}{\sqrt{2}}\cdot \lim _{x\to \:0+}\left(\frac{\sqrt{1-\cos \left(2x\right)}}{x}\right)\\\mathrm{Apply\:L’Hospital’s\:Rule}\\=\frac{1}{\sqrt{2}}\cdot \lim _{x\to \:0+}\left(\frac{\frac{\sin \left(2x\right)}{\sqrt{1-\cos \left(2x\right)}}}{1}\right)\\=\frac{1}{\sqrt{2}}\cdot \lim _{x\to \:0+}\left(\frac{\sin \left(2x\right)}{\sqrt{1-\cos \left(2x\right)}}\right)\\\mathrm{Apply\:L’Hospital’s\:Rule}\\=\frac{1}{\sqrt{2}}\cdot \lim _{x\to \:0+}\left(\frac{\cos \left(2x\right)\cdot \:2}{\frac{\sin \left(2x\right)}{\sqrt{1-\cos \left(2x\right)}}}\right)\\\mathrm{Simplify\:}\frac{\cos \left(2x\right)\cdot \:2}{\frac{\sin \left(2x\right)}{\sqrt{1-\cos \left(2x\right)}}}\\=\frac{1}{\sqrt{2}}\cdot \lim _{x\to \:0+}\left(2\cot \left(2x\right)\sqrt{1-\cos \left(2x\right)}\right)\\=\frac{1}{\sqrt{2}}\cdot \:2\cdot \lim _{x\to \:0+}\left(\cot \left(2x\right)\sqrt{1-\cos \left(2x\right)}\right)\\\mathrm{Multiply\:by\:the\:conjugate\:of}\:1-\cos \left(2x\right)\\=\frac{\left(1-\cos \left(2x\right)\right)\left(1+\cos \left(2x\right)\right)}{1+\cos \left(2x\right)}\\\mathrm{Expand}\:\left(1-\cos \left(2x\right)\right)\left(1+\cos \left(2x\right)\right)\\=\frac{\sin ^2\left(2x\right)}{1+\cos \left(2x\right)}\\=\frac{1}{\sqrt{2}}\cdot \:2\cdot \lim _{x\to \:0+}\left(\cot \left(2x\right)\sqrt{\frac{\sin ^2\left(2x\right)}{1+\cos \left(2x\right)}}\right)\\=\frac{1}{\sqrt{2}}\cdot \:2\cdot \lim _{x\to \:0+}\left(\cos \left(2x\right)\sqrt{\frac{1}{\cos \left(2x\right)+1}}\right)\\\mathrm{Plug\:in\:the\:value}\:x=0\\=\frac{1}{\sqrt{2}}\cdot \:2\cos \left(2\cdot \:0\right)\sqrt{\frac{1}{\cos \left(2\cdot \:0\right)+1}}\\\mathrm{Simplify\:} \frac{1}{\sqrt{2}}\cdot \:2\cos \left(2\cdot \:0\right)\sqrt{\frac{1}{\cos \left(2\cdot \:0\right)+1}}\\=1\\\mathrm{Similarly\:} \lim _{x\to \:0-}\left(\frac{\sqrt{1-\cos \left(2x\right)}}{\sqrt{2}x}\right)=-1\\=\mathrm{diverges}$