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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!

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Word of advice: in most cases you might like to try out Taylor series before turning to LHR, which generally mess things up and should serve as the last resort rather than the first trial. – Vim Mar 28 at 2:07
up vote 3 down vote accepted

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 $$\lim_{x\to 0^+}\frac{\sqrt{e^x-1} - \sqrt x}{x} = \lim_{x\to 0^+}\frac{e^x-1 - x}{x\sqrt{e^x-1} + x\sqrt x } = 0.$$

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Excellent answer!!! – Konstantinos Gaitanas Mar 27 at 19:12

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 $x \to 0$.

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