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  • Help me answer: - Limit , continuity and differentiability - JEE Main-5

\lim_{x\rightarrow 0}\frac{log_{e}(1+2x)}{x}  equals 

  • Option 1)

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  • Option 2)

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  • Option 3)

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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}(1+2x)}{2x}*2= 2*1=2

 

 

 

 


Option 1)

1

Option 2)

2

Option 3)

3

Option 4)

4

Posted by

Himanshu

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