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  • Can someone help me with this, - Limit , continuity and differentiability - JEE Main-6

\lim_{x\rightarrow 0} \frac{log _{e}(\frac{1+2x}{1-2x})}{x} equals 

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  • Option 2)

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  • Option 4)

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Answers (1)

best_answer

As we have learned

Evaluation of Logarithmic Limits -

\lim_{x\rightarrow 0}\:\:\:\frac{log_{e}(1+x)}{x}=1

\Rightarrow \lim_{x\rightarrow 0}\:\:\:\frac{log_{e}(1+Kx)}{x}\neq 1

but\lim_{x\rightarrow 0}\:\:\:\frac{log_{e}(1+Kx)}{x}\times \frac{K}{K}

\therefore \:\:\:K\lim_{x\rightarrow 0}\:\:\:\frac{log_{e}(1+Kx)}{Kx}

=K\times 1=K

\Rightarrow \lim_{x\rightarrow 0}\:\:\:\frac{log_{e}(1-x)}{x}=-1

- wherein

\frac{1+x}{x}

x\:must\:be\:same

 

 \lim_{x\rightarrow 0} \frac{log _{e}\left ( \frac{1+2x}{1-2x} \right )}{x}=\lim_{x\rightarrow 0}\frac{log _{e}(1+2x)-log _{e}(1-2x)}{x}

=\lim_{x\rightarrow 0}\frac{log_{e}(1+2x)}{x}+\lim_{x\rightarrow 0}\frac{log _{e}(1-2x)}{-x}

2\lim_{x\rightarrow 0} \frac{log _{e}(1+2x)}{2x}+ 2\lim_{x\rightarrow 0}\frac{log_{e}(1-2x)}{-2x}

2(1)+2(1)=4

 

 

 

 


Option 1)

1

Option 2)

2

Option 3)

3

Option 4)

4

Posted by

Himanshu

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