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32 votes
32 votes

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

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

Best answer
46 votes
46 votes

Apply an exponential of a logarithm to the expression.

$\quad\lim\limits_{x\to \infty}(x^{2}+1)^{e^{-x}}=\lim\limits_{x\to \infty} \exp \left(\log ((x^{2}+1)^{e^{-x}})\right)$

$=\lim\limits_{x\to \infty} \exp \left(\dfrac{\log(x^{2}+1)}{e^{x}}\right)$


Since the exponential function is continuous, we may factor it out of the limit.

$\quad\lim\limits_{x\to \infty} \exp \left(\dfrac{\log(x^{2}+1)}{e^{x}}\right)$

$=\exp \left(\lim\limits_{x\to \infty} \dfrac{\log(x^{2}+1)}{e^{x}}\right)$


The numerator of $e^{-x} \log (x^{2}+1)$  grows asymptotically slower than its denominator as $x$ approaches $\infty$.

Since $\log (x^{2}+1)$ grows asymptotically slower than $e^{x}$ as $x$ approaches $\infty$, $\lim\limits_{x\to \infty}e^{-x} \log (x^{2}+1)=0 : e^{0}.$


Evaluate $e^{0}$.

$e^{0}=1$:

Answer: $1.$

Correct Answer: $C$

edited by
81 votes
81 votes

$\lim_{x \to \infty}(1+x^2)^{e^-x}=y(let)$

Take log

$log(y)=e^{-x}.log(1+x^2)=lim_{x \to \infty} \frac{log(1+x^2)}{e^x}=\frac{\infty}{\infty}form$

Apply L Hospital rule

$lim_{x \to \infty} \frac{\frac{2x}{1+x^2}}{e^x}=\frac{2x}{e^x(1+x^2)}$

Differentiate one-more time

$\lim_{x \to \infty} \frac{2}{e^x(1+x^2)+2xe^x}=0$

$log(y)=0 \rightarrow y=1$

Answer : (1)

Tip: For power functions it always better to take log and solve.

13 votes
13 votes

Here the base of the log is assumed to be e(Natural log).

Hence Y=e0 = 1 is ans.(The last step)

8 votes
8 votes

I think this would help you to solve this type of question in much easier way 

Answer:

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