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:

Graphical Representation of Limits

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] limx0[sinxx]=12] limx0[tanxx]=13] limx(1+1x)x=e4] limx0ln(1+x)x=15] limx0loga(1+x)x=1lna6] limx0ax1x=lna7] limx1xm1x1=m8] limxaxmamxa=mam19] limx0sin1xx=limx0tan1xx=11] \ \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

Solved Practice Examples on Limits

Solution:

Consider Practice Examples on Limits

Examples on Limits

Solved Practice Examples on Limits for JEE

Solved Practice Examples on Limits for IIT JEE

Example 3: Evaluate

Solve Limit Problems

Solution:

Limit Problems 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:

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, 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: limx0(1cos2x2x)\lim _{x\to 0}\left(\frac{\sqrt{1-cos2x}}{\sqrt{2}x}\right)

Solution: 

Iflimxaf(x)limxa+f(x)thenthelimitdoesnotexist\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}

limx0+(1cos(2x)2x)limxa[cf(x)]=climxaf(x)=12limx0+(1cos(2x)x)ApplyLHospitalsRule=12limx0+(sin(2x)1cos(2x)1)=12limx0+(sin(2x)1cos(2x))ApplyLHospitalsRule=12limx0+(cos(2x)2sin(2x)1cos(2x))Simplifycos(2x)2sin(2x)1cos(2x)=12limx0+(2cot(2x)1cos(2x))=122limx0+(cot(2x)1cos(2x))Multiplybytheconjugateof1cos(2x)=(1cos(2x))(1+cos(2x))1+cos(2x)Expand(1cos(2x))(1+cos(2x))=sin2(2x)1+cos(2x)=122limx0+(cot(2x)sin2(2x)1+cos(2x))=122limx0+(cos(2x)1cos(2x)+1)Pluginthevaluex=0=122cos(20)1cos(20)+1Simplify122cos(20)1cos(20)+1=1Similarlylimx0(1cos(2x)2x)=1=diverges\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}