GATE CS 2015 Set 1 Q.14 (CS Q.4): Calculus

$\lim_{x \to \infty} x^{1/x}$ is:

$\infty$

$0$

$1$

Not defined

Official answer and solution

Solution

Transform the $x^{1/x}$ form into an exponential form and apply L'Hopital's rule as $x \to \infty$ .

Let $L = \lim_{x \to \infty} x^{1/x}$ , which is of the indeterminate form $\infty^0$ .

Taking the natural logarithm of both sides:

\ln L = \lim_{x \to \infty} \ln(x^{1/x}) = \lim_{x \to \infty} \frac{\ln x}{x}.

As $x \to \infty$ , $\frac{\ln x}{x}$ is of the form $\frac{\infty}{\infty}$ . Applying L'Hopital's Rule:

\lim_{x \to \infty} \frac{\frac{d}{dx}(\ln x)}{\frac{d}{dx}(x)} = \lim_{x \to \infty} \frac{1/x}{1} = \lim_{x \to \infty} \frac{1}{x} = 0.

Since $\ln L = 0$ , we exponentiate to obtain:

L = e^0 = 1.

This matches option C.

Final answer

Common trap

Concluding that $x^{1/x} \to 0$ because $1/x \to 0$ , forgetting that $x \to \infty$ makes $e^{\ln x / x} \to e^0 = 1$ .

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