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Limits of the form (1 power infinity) - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Limits of the form (1 power infinity) is considered one the most difficult concept.

  • 206 Questions around this concept.

Solve by difficulty

\lim_{x\rightarrow \infty }\left ( \frac{x^{2}+5x+3}{x^{2}+x+3} \right )^{\frac{1}{x}}

If \lim_{x\rightarrow \infty }\left ( 1+\frac{a}{x}+\frac{b}{x^{2}} \right )^{2x}= e^{2}, then the values of a and b are

Let p= \lim_{x\rightarrow 0+}\left ( 1+\tan ^{2} \sqrt{x}\right )^{\frac{1}{2x}}then  log p
is equal to :

limx0(1+x)1/xexequals

Concepts Covered - 1

Limits of the form (1 power infinity)

Limits of the form 1 (1 power infinity)

To find the limit of the form 1, we will use the following results
If limxaf(x)=limxag(x)=0
Then,

limxa[1+f(x)]1g(x)=elimxaf(x)g(x)
Or
When, limxaf(x)=1 and limxag(x)=
Then,

limxa[f(x)]g(x)=limxa[1+(f(x)1)]g(x)=elimxa(f(x)1)g(x)

Proof: 

Let L=limxa[1+f(x)]1g(x)
Taking logs of both sides

log(L)=log(limxa[1+f(x)]1g(x))log(L)=limxa1g(x)log(1+f(x))log(L)=limxa1g(x)(log(1+f(x))f(x))f(x)
As f(x) is tending to 0 , so

log(L)=limxaf(x)g(x)L=elimxaf(x)g(x)2

Some particular cases  

(a) limx0(1+x)1x=e
(b) limx(1+1x)x=e
(c) limx0(1+cx)1x=ec
(d) limx(1+cx)x=ec

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Limits of the form (1 power infinity)

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Reference Books

Limits of the form (1 power infinity)

Mathematics for Joint Entrance Examination JEE (Advanced) : Calculus

Page No. : 2.26

Line : 32

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