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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.

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QuestionMEDIUM

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

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

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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:

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

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

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\mathrm{\mathrm{\underset{x \rightarrow \/\infty }{\operatorname{Lim}}}\, \, \frac{e^{x^2}-1}{e^{x^2}+1}=1}

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\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}}=}     ______________

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\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.
 

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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

Study other Related Concepts

Exponential and Logarithmic Limits Current Topic

Limit
Left-Hand Limits and Right-Hand Limits
Algebra of Limits
Limit of Indeterminate Form and Algebraic limit
Limit of Algebraic function
Limit Using Expansion
Sandwich Theorem
Limits of Trigonometric Functions
Exponential and Logarithmic Limits
Limits of the form (1 power infinity)

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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

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