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  • Confused! kindly explain, - Limit , continuity and differentiability - JEE Main-9

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

  • Option 1)

    1/e

  • Option 2)

    1/e^{2}

  • Option 3)

    1/e^{3}

  • Option 4)

    1

 

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}

 

 

-

 

its again of form 1^{\infty }  so 

limit = e^{\lim_{x\rightarrow 0}\left ( \frac{x+1}{x+2}-1 \right )\cdot x}= e^{\lim_{x\rightarrow \infty }\frac{-x}{x+2}}

\frac{-x}{x+2}  has degree of numerator and denominator 

same , so \lim_{x\rightarrow \infty }\frac{-x}{x+2}=-1/1=-1 

\therefore find limit = e^{-1}= 1/e  

 

 

 

 

 


Option 1)

1/e

Option 2)

1/e^{2}

Option 3)

1/e^{3}

Option 4)

1

Posted by

Himanshu

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