GATE CS 2022 Q.34: Calculus

The value of the following limit is _____________.

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

Your answer

Official answer and solution

Solution

Evaluate the indeterminate form 0/0 by applying L'Hopital's Rule.

Direct substitution of $x = 0$ gives the indeterminate form $\frac{0}{1 - e^0} = \frac{0}{0}$ .

Applying L'Hopital's rule by differentiating numerator and denominator with respect to $x$ :

\lim_{x \to 0^+} \frac{\frac{d}{dx}(x)}{\frac{d}{dx}(1 - e^{2x})} = \lim_{x \to 0^+} \frac{1}{-2e^{2x}}

Evaluating the limit at $x = 0$ :

\frac{1}{-2e^0} = -\frac{1}{2} = -0.5

Final answer

-0.5

Common trap

Missing the negative sign from the denominator derivative $\frac{d}{dx}(1 - e^{2x}) = -2e^{2x}$ .

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

GATE questions are the work of the organising institutes. This site is independent and is not affiliated with them.