GATE CS 2022 Q24 — Calculus: Limits, Continuity, Differentiability

To evaluate the limit:

$$\lim_{x \rightarrow 0^{+}} \frac{\sqrt{x}}{1-e^{2\sqrt{x}}}$$

Let us substitute $t = \sqrt{x}$. As $x \rightarrow 0^{+}$, we have $t \rightarrow 0^{+}$.

Substituting $t$ into the limit expression gives:

$$\lim_{t \rightarrow 0^{+}} \frac{t}{1 - e^{2t}}$$

This is an indeterminate form of the type $\frac{0}{0}$. We can evaluate this using L'Hôpital's Rule by differentiating the numerator and the denominator with respect to $t$:

$$\lim_{t \rightarrow 0^{+}} \frac{\frac{d}{dt}(t)}{\frac{d}{dt}(1 - e^{2t})} = \lim_{t \rightarrow 0^{+}} \frac{1}{-2e^{2t}}$$

Now, substituting $t = 0$:

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

Thus, the value of the limit is -0.5.