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# Exponential and Logarithmic Limits - Practice Questions & MCQ

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

## Quick Facts

• Exponential Limits is considered one the most difficult concept.

• Logarithmic  Limits is considered one of the most asked concept.

• 271 Questions around this concept.

## Solve by difficulty

Choose the right expansion which is represented by the limit $\lim _{x \rightarrow 0} \frac{a^{x}-1}{x}$.

Choose the right expansion which is represented by the limit  $\lim _{x \rightarrow 0} \frac{a^{x}-1}{x}$.

If $\lim _{x \rightarrow 0} \frac{(729)^x-(243)^x-(81)^x+9^x+3^x-1}{x^3}=\lambda(\ln 3)^3$  $\lambda$ is equal to:

Evaluate $\mathrm{\lim _{x \rightarrow 0} \frac{\mathrm{e}^{\mathrm{x}}-\mathrm{e}^{-x}-2 x}{\sin ^3 x}}$

$\mathrm{ \lim _{x \rightarrow 0}\left\{\frac{e^x-e^{\sin x}}{(x-\sin x)}\right\} \text { is equal to }}$

$\mathrm{ \lim _{x \rightarrow \infty}\left[e^{\sqrt{\left(x^4+1\right)}}-e^{\left(x^2+1\right)}\right]}$

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$\mathrm{\lim_{x\rightarrow 0}\frac{1}{10} \frac{1-e^{-j5 x}}{1-e^{-j x}}=}$     ______________

$\mathrm{\lim _{x \rightarrow 0} \frac{e^{x^2}-\cos x}{x^2}}$ is equal to

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The value of  $\mathrm{\lim _{x \rightarrow 0} \frac{e^{1 / x}-1}{e^{1 / x}+1}, x \neq 0}$ is.

## Concepts Covered - 2

Exponential Limits

Exponential Limits

In order to solve the limit of the function involving exponential function, we use the following standard results

$\\\text{(i)}\;\;\;\mathbf{\lim _{x \rightarrow 0} \frac{a^{x}-1}{x}=\log _{e} a}$

Proof:

$\\ \lim _{x \rightarrow 0} \frac{a^{x}-1}{x}=\lim _{x \rightarrow 0} \frac{\left(1+\frac{x(\log a)}{1 !}+\frac{x^{2}(\log a)^{2}}{2 !}+\cdots\right)-1}{x} \\\\\text{[using Taylor series expansion of }a^x]\\\\\text{}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\lim _{x \rightarrow 0}\left(\frac{\log a}{1 !}+\frac{x(\log a)^{2}}{2 !}+\cdots\right)\\\\\text{}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\log_ea$

$\\\text{(ii)}\;\;\;\mathbf{\lim _{x \rightarrow 0} \frac{e^{x}-1}{x}=1}$

In General, if $\lim_{x\rightarrow a}f(x)=0$ , then we have

$\\\text{(a)}\;\;\;\lim _{x \rightarrow a} \frac{a^{f(x)}-1}{f(x)}=\log_ea\\\\\text{(b)}\;\;\;\lim _{x \rightarrow a} \frac{e^{f(x)}-1}{f(x)}=\log_ee=1$

Logarithmic  Limits

Logarithmic  Limits

To evaluate the Logarithmic limit we use the following results:

$\\\mathbf{\lim _{x \rightarrow 0} \frac{\log_e (1+x)}{x}=1}$

Proof:

$\\\lim _{x \rightarrow 0} \frac{\log_e (1+x)}{x}=\lim _{x \rightarrow 0} \frac{x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\cdots}{x}\\\\\text{[using Taylor series expansion of }\log_e(1+x)]\\\\\text{}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\lim _{x \rightarrow 0}\left(1-\frac{x}{2}+\frac{x^{2}}{3}-\cdots\right)\\\\\text{}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=1$

In General, if $\lim_{x\rightarrow a}f(x)=0$ , then we have $\lim _{x \rightarrow a} \frac{\log_e (1+f(x))}{f(x)}=1$

## Study it with Videos

Exponential Limits
Logarithmic  Limits

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## Books

### Reference Books

#### Exponential Limits

Mathematics for Joint Entrance Examination JEE (Advanced) : Calculus

Page No. : 2.23

Line : 12

#### Logarithmic  Limits

Mathematics for Joint Entrance Examination JEE (Advanced) : Calculus

Page No. : 2.23

Line : 21