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  • Help me answer: - Limit , continuity and differentiability - JEE Main-6

\lim_{x\rightarrow 0}\left ( 1+2x \right )^{1/ \sin x}   equals 

  • Option 1)

    1

  • Option 2)

    e

  • Option 3)

    e^{2}

  • Option 4)

    e^{3}

 

Answers (1)

best_answer

As we have learned

Result of 1 to the power of infinity Form -

\lim_{x\rightarrow \infty}\left(1+\frac{1}{x} \right )^{x}=e

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

\lim_{x\rightarrow 0}\:\;\:(1+\lambda x)^{\frac{1}{x}}=e^{\lambda }

\lim_{x\rightarrow \infty}\left(1+\frac{\lambda}{x} \right )^{x}=e^{\lambda}

 

 

-

 

 Again to form 1^{\infty } , so limit = e^{\lim_{x\rightarrow 0}(2x)\cdot 1/\sin x}

= e^{2\lim_{x\rightarrow 0}1/\sin x}=e^{2*1}=e^{2}

 

 

 

 


Option 1)

1

Option 2)

e

Option 3)

e^{2}

Option 4)

e^{3}

Posted by

Himanshu

View full answer

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