Convergence of $\int_0^1\frac{e^{\sqrt x}-1}{x}dx$ 
Determine if the following integral will converge.
  $$\int_0^1\frac{e^{\sqrt x}-1}{x}dx$$

My approach was something like this. I made the assumption that $e^{\sqrt{x}} − 1 ≈ x$ and then followed like this:
$$\frac{e^{\sqrt x}-1}{x} \overset{\mathrm{L'Hosp.}}{=}  \frac{e^\sqrt{x}}{2\sqrt{x}\over1}$$
But this doesn't seem to be getting anywhere. Any tips for this?
 A: You can use the comparison criteria for improper integrals. Noting that
$$
\lim_{x\to 0}\frac{\frac{e^{\sqrt{x}}-1}{x}}{\frac{1}{\sqrt{x}}}  = 1,
$$
you  can establish that the given integral has the same nature as $\int_0^1 \frac{1}{\sqrt{x}} dx$, which is convergent.
A: Since
$1+x
\le e^x
\le 1+x+x^2
$
for
$0 \le x \lt 1$,
$\int_0^1\frac{e^{\sqrt x}-1}{x}dx
\le \int_0^1\frac{\sqrt{x}+x}{x}dx
= \int_0^1(\frac1{\sqrt{x}}+1)dx
=(2\sqrt{x}+x)|_0^1
=3
$.
Also
$\int_0^1\frac{e^{\sqrt x}-1}{x}dx
\ge \int_0^1\frac{\sqrt{x}}{x}dx
=(2\sqrt{x})|_0^1
=2
$.
A: Rather than $\dfrac{e^\sqrt{x}}{\left( 2\sqrt{x}\over1 \right)}$ you should have $\dfrac{\left( \dfrac{e^{\sqrt x}}{2\sqrt x} \right)}{1}$ in your application of L'Hopital's rule.
Once you've got that, you can do this:
$$
\lim_{x\,\downarrow\,0} \frac{e^{\sqrt x}}{2\sqrt x} = \lim_{u\,\downarrow\,0} \frac{e^u} {2u} = \text{etc.} \\ \text{(since $u=\sqrt x\to0$ as $x\to0$)}
$$
But you can do it without L'Hopital's rule, as follows:
\begin{align}
\lim_{x\,\downarrow\,0} \frac{e^{\sqrt x}-e^{\sqrt0}} {x-0} = {} & f'(0) \quad \text{where } f(x) = e^{\sqrt x} \\[8pt]
= {} & e^{\sqrt x} \cdot \frac d {dx} \sqrt x \quad \text{etc.}
\end{align}
A: Let $y=\sqrt{x}$ so that $x=y^2$ and $dx=2y\,dy.$
$$\int_0^1 \frac{e^{\sqrt{x}}-1}{x} dx = 2\int_0^1 \frac{e^y-1}{y}\,dy.$$
Because the denominator goes to zero, we investigate the limit of $\displaystyle f(y)=\frac{e^y-1}{y}.$
$$\lim_{y\rightarrow 0} f(y) = \lim_{y\rightarrow 0} \frac{e^y-1}{y}=\lim_{y\rightarrow 0} \frac{e^y}{1}=1, \quad \textrm{by L'Hopital's rule}.$$
Also, as @martycohen pointed out, $e^y\le 1+y+y^2$ on the interval $[0,1]$, so 
$$2\int_0^1 \frac{e^y-1}{y}\,dy\le2\int_0^1 \frac{y+y^2}{y}\,dy=2\int_0^1 {(1+y)}\,dy=3.$$
