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The value of  \mathrm{\lim _{x \rightarrow \infty}\left(1+x^2\right)^{e^{-x}}} is

Option: 1

0


Option: 2

\frac{1}{2}


Option: 3

1


Option: 4

\infty


Answers (1)

best_answer

Let               \mathrm{K=\lim _{x \rightarrow \infty}\left(1+x^2\right)^{e^{-x}} \quad\left(\infty^0 \text { Form }\right)}

Taking logarithms

                      \mathrm{\begin{aligned} \log K & =\lim _{x \rightarrow \infty} e^{-x} \cdot \log \left(1+x^2\right)(0 \cdot \infty \text { form }) \\ & =\lim _{x \rightarrow \infty} \frac{\log \left(1+x^2\right)}{e^x}\left(\frac{\infty}{\infty} \text { form }\right) \end{aligned}}

                                  \mathrm{=\lim _{x \rightarrow \infty} \frac{\left(\frac{2 x}{1+x^2}\right)}{e^x}}

                                                          (By L Hospital's Rule)

                                \mathrm{=\lim _{x \rightarrow \infty}\left[\frac{2 x}{\left(1+x^2\right) e^x}\right]\left(\frac{\infty}{\infty} \text { form }\right)}

                                                             (By L Hospital's Rule)

                                \frac{2}{\infty}=0

So ,                         \mathrm{K=e^0=1}

Posted by

himanshu.meshram

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