GATE 2022 CS – Question 34
The value of the following limit is _____________.
$\lim_{x\to0^+}\frac{\sqrt{x}}{1-e^{2\sqrt{x}}}$
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Correct answer: -0.5
Explanation
Let $t=\sqrt{x}$. Then $\frac{t}{1-e^{2t}}\to\frac{t}{-2t}=-\frac12$, since $1-e^{2t}\approx-2t$.