Lim x->0 (e^sqrt(x) -2)/x
solve it.
Thank
I solved it but I've got 2 answers and I'm not sure. Sorry about my english. I'm not a native speaker .
perhaps you were a bit over-eager to use l'hopital?
we can use l'hopital once, to get:
$\displaystyle \lim_{x \to 0^+} \frac{e^{\sqrt{x}} - 1}{x} = \lim_{x \to 0^+} \frac{e^{\sqrt{x}}}{2\sqrt{x}}$
but in the latter limit, the numerator goes to 1, while the denominator goes to 0, so the ratio is unbounded.
If you allow complex values, then it is alright to use a two-sided limit but then if you are allowing such things, you might as well consider the limit on the complex plane. This might be a bit of a tangent but here we go. Let
$\displaystyle f(z) = \frac{e^{\sqrt{z}}-1}{z}$
for complex values $\displaystyle z$. We'll show that $\displaystyle f(z)$ is analytic. The condition with the Cauchy-Riemann equations is equivalent to saying that
$\displaystyle \frac{\partial f}{\partial \overline{z}} = \frac{1}{2}\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)f = 0$
for $\displaystyle z = x + i y$
If you calculate this, it is indeed equal to zero. So the function is analytic and we can take the limit as $\displaystyle z \to 0$. The limit happens to be real and unbounded in the positive direction.