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GATE | GATE-CS-2015 (Set 3) | Question 65

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\\ The \hspace{2 mm}value \hspace{2 mm} of \hspace{2 mm} \lim_{x\to\infty }(1 + x^{2})e^{-x} is
(A) 0
(B) 1/2
(C) 1
(D) ∞


Answer: (A)

Explanation: This can be solved using L’Hôpital’s rule that uses derivatives to help evaluate limits involving indeterminate forms.

Since \lim_{x \to c}f(x)=\lim_{x \to c}g(x)=\infty, and \lim_{x\to c}\frac{f'(x)}{g'(x)} exists

We get
\lim_{x\to c}\frac{f(x)}{g(x)} = \lim_{x\to c}\frac{f'(x)}{g'(x)}.

\lim_{x\to \infty}\frac{1 + x^2}{e^x} = \lim_{x\to \infty}\frac{2x}{e^x} = \lim_{x\to \infty}\frac{2}{e^x} = 0


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Last Updated : 15 Feb, 2018
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