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GATE CSE 2015 | Set 3 | Question: 9

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The value of $\displaystyle \lim_{x \rightarrow \infty} (1+x^2)^{e^{-x}}$ is

  1. $0$
  2. $\frac{1}{2}$
  3. $1$
  4. $\infty$

8 Answers

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take e^(-x) =t    also, x = -lnt then x^2 = 2lnt 

the when x goes to infinity t goes to zero 

then limit becomes t->0 (1+ 2lnt) ^ t 

this becomes 1^0 = 1

ans is 1

 

 

 

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