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The value of the following limit is ________________.

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

Given,

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

Using L'hopital rule

$\displaystyle{}\lim_{x\to0} \frac{\frac{1}{2\sqrt(x)}}{-e^{2\sqrt(x)} \times \frac{2}{2\sqrt(x)}} = \lim_{x\to0} \frac{1}{-2e^{2\sqrt(x)}} = -\frac{1}{2} = -0.5$

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Thanks.

$\displaystyle \lim_{x \to 0+} \frac{\sqrt{x}}{1-e^{2\sqrt{x}}}\; (\frac{0}{0}\,form)$

$\text{Using L'Hôpital's rule}$

$\displaystyle \lim_{x \to 0+} \frac{\frac{1}{2\sqrt{x}}}{0-e^{2\sqrt{x}}.\frac{2}{2\sqrt{x}}} =\lim_{x \to 0+} -\frac{1}{2.e^{2\sqrt{x}}} =-\frac{1}{2} =-0.5$
Using, $e^x = \Sigma_{n=0}^{\infty}\frac{x^n}{n!}$ (Taylor Series Expansion for $e^x$ near $x=0$)

$\lim_{x \rightarrow 0^+} \frac{\sqrt{x}}{1-\left(1 + \frac{2\sqrt{x}}{1!} + \frac{(2\sqrt{x})^2}{2!} +... \right )} = \lim_{x \rightarrow 0^+} \frac{\sqrt{x}}{-\left( \frac{2\sqrt{x}}{1!} + \frac{(2\sqrt{x})^2}{2!} +... \right )}$

$= \lim_{x \rightarrow 0^+} \frac{-1}{\left( \frac{2}{1!} + \frac{2^2\sqrt{x}}{2!} +... \right )}, x \neq 0$

$= \frac{-1}{2}$