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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.

  • 195 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 :

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 \lim_{x\rightarrow a}\;f(x)=\lim_{x\rightarrow a}\;g(x)=0

Then,

\lim_{x\rightarrow a}\;\left [1+f(x) \right ]^{\frac{1}{g(x)}}=e^{\lim_{\;x\rightarrow a}\frac{f(x)}{g(x)}} .

Or

When,  \lim_{x\rightarrow a}\;f(x)=1 and \lim_{x\rightarrow a}\;g(x)=\infty

Then, 

\\\lim_{x\rightarrow a}\;\left [f(x) \right ]^{{g(x)}}=\lim_{x\rightarrow a}\;\left [1+(f(x)-1) \right ]^{{g(x)}}\\\\\text{}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=e^{\lim_{\;x\rightarrow a}{\left (f(x)-1 \right )}{g(x)}}

 

Proof: 

Let L=\lim_{x\rightarrow a}\;\left [1+f(x) \right ]^{\frac{1}{g(x)}}

Taking log of both sides

log(L)=log\left( \lim_{x\rightarrow a}\;\left [1+f(x) \right ]^{\frac{1}{g(x)}}\right)

log(L)=\lim_{x\rightarrow a}\frac{1}{g(x)}log\left( \;1+f(x)\right)

log(L)=\lim_{x\rightarrow a}\frac{1}{g(x)}\left(\frac{log\left( \;1+f(x)\right)}{f(x)}\right).f(x)

As f(x) is tending to 0, so

log(L)=\lim_{x\rightarrow a}\frac{f(x)}{g(x)}

L=e^{\lim_{x\rightarrow a}\frac{f(x)}{g(x)}}

 

 

Some particular cases  

\\\text{(a)}\;\;\;\lim_{x\rightarrow 0}\;(1+x)^{\frac{1}{x}}=e\\\\\text{(b)}\;\;\;\lim_{x\rightarrow \infty}\;\left (1+\frac{1}{x} \right )^{x}=e\\\\\text{(c)}\;\;\;\lim_{x\rightarrow 0}\;(1+cx)^{\frac{1}{x}}=e^c\\\\\text{(d)}\;\;\;\lim_{x\rightarrow \infty}\;\left (1+\frac{c}{x} \right )^{x}=e^c

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Limits of the form (1 power infinity)

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