Trying to compute a limit Hello all I would like to compute the limit
$$\lim_{x\to\ 0^{+}} \frac{\sqrt{e^x-1}-\sqrt{x}}{x}$$
I know from Wolfram that the value of the limit is $0$ but I would like to know why.
I tried to apply L' Hospital's rule but things are getting (much) worse.
Any hints would be appreciated.
Thanks in advance!
 A: For $\lim_{x\to\ 0^{+}} \frac{\sqrt{e^x-1}-\sqrt{x}}{x}$ expand $e^x$ as $1 + x + \frac{x^2}{2}$ then $\sqrt{e^x-1} \approx \sqrt{x + x^2/2}$. Take out $\sqrt{x}$ as a common factor so that $\sqrt{x + x^2/2} = \sqrt{x} \, \sqrt{1 + x/2}$. Expand the second square root by the Binomial Theorem, so that eventually you have
 $\frac{\sqrt{e^x-1}-\sqrt{x}}{x} \approx \frac{\sqrt{x}(1 + x/4) - \sqrt{x}}{x}$. This yields $\frac{\sqrt{e^x-1}-\sqrt{x}}{x} \approx \frac{\sqrt{x}x/4}{x}$, finally giving $\sqrt{x}/4$ which tends to zero as $x \to 0$.
A: Multiplying on top and bottom by $\sqrt{e^x-1} + \sqrt x$, we see $$ \frac{\sqrt{e^x-1} - \sqrt x}{x} = \frac{e^x-1 - x}{x\sqrt{e^x-1} + x\sqrt x }.$$ Now $e^x \ge 1+x$ so we see \begin{equation}0 \le \frac{e^x-1 - x}{x\sqrt{e^x-1} + x\sqrt x } \le \frac{e^x -1 - x}{x \sqrt x} = \frac{e^x -1 - x}{x^{3/2}}. \,\,\,\,\,\,\,\,\,\,\,\,\, (*)\end{equation} Now $$\lim_{x \to 0^+} \frac{e^x -1 - x}{x^{3/2}} = \lim_{x \to 0^+} \frac{e^x -1}{\tfrac 3 2 x^{1/2}} = \lim_{x \to 0^+} \frac{e^x}{\tfrac 3 4x^{-1/2}} = \lim_{x \to 0^+} \tfrac 4 3 \sqrt x e^x = 0$$ by l'Hopital's rule. Thus by equation $(*)$ and the squeeze theorem, we get that $$\lim_{x\to 0^+}\frac{\sqrt{e^x-1} - \sqrt x}{x} = \lim_{x\to 0^+}\frac{e^x-1 - x}{x\sqrt{e^x-1} + x\sqrt x } = 0.$$
