GATE CS 2015 Set 3 Q.19 (CS Q.9): Calculus

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

$0$

$\frac{1}{2}$

$1$

$\infty$

Official answer and solution

Official answer: C
From: Official GATE 2015 answer key

Solution

Apply logarithmic transformation and L'Hopital's rule to evaluate the limit at infinity.

Let $L = \lim_{x \to \infty} (1 + x^2)^{e^{-x}}$ .
Taking the natural logarithm of both sides:

\ln L = \lim_{x \to \infty} \ln\left((1 + x^2)^{e^{-x}}\right) = \lim_{x \to \infty} e^{-x} \ln(1 + x^2) = \lim_{x \to \infty} \frac{\ln(1 + x^2)}{e^x}

As $x \to \infty$ , $\ln(1 + x^2) \to \infty$ and $e^x \to \infty$ , which is an indeterminate form of type $\frac{\infty}{\infty}$ .
Applying L'Hopital's rule:

\lim_{x \to \infty} \frac{\frac{d}{dx}[\ln(1 + x^2)]}{\frac{d}{dx}[e^x]} = \lim_{x \to \infty} \frac{\frac{2x}{1 + x^2}}{e^x} = \lim_{x \to \infty} \frac{2x}{(1 + x^2) e^x}

As $x \to \infty$ :

\frac{2x}{(1 + x^2) e^x} = \frac{2}{x \left(1 + \frac{1}{x^2}\right) e^x} \to 0

Therefore:

\ln L = 0 \implies L = e^0 = 1

Hence, the value of the limit is 1 (Option C).

Final answer

Common trap

Evaluating limit as x -> 0 instead of x -> infinity, or wrongly assuming that the base (1 + x^2) -> infinity forces the limit to infinity without accounting for the rapidly decaying exponent e^(-x).

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2015 Computer Science and Information Technology (CS) Session 3
Question: 19 of 65, printed as CS Q.9
Section: Computer Science and Information Technology

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