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$\;\lim \limits_{x \to \infty}\large\frac{x^{2}}{e^{x}}$ is

\[\begin{array}{1 1}(1)2&(2)0\\(3)\infty&(4)1\end{array}\]

1 Answer

$\lim \limits_{x \to \infty } \large\frac{x^2}{e^x} =\frac{\infty^2}{e^x}=\frac{\infty}{\infty}$
By L Hospital Rule
$\lim \limits_{x \to \infty } \large\frac{x^2}{e^x} $$=\lim \limits_{x \to \infty } \large\frac{2x}{e^x}$
$\qquad=\lim \limits_{x \to \infty } \large\frac{x^2}{e^x}$
$\qquad= \large\frac{2}{e^x}=\frac{2}{\infty}=$$0$
Hence 2 is the correct answer.
answered May 19, 2014 by meena.p
 

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