Thanks.

Dark Mode

1,404 views

1 vote

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}$

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

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

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