Lim x->0 (e^sqrt(x) -2)/x

solve it.

Thank :D

I solved it but I've got 2 answers and I'm not sure. Sorry about my english. I'm not a native speaker .

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- Aug 19th 2012, 08:59 AMTiffanyIt's limit.
Lim x->0 (e^sqrt(x) -2)/x

solve it.

Thank :D

I solved it but I've got 2 answers and I'm not sure. Sorry about my english. I'm not a native speaker . - Aug 19th 2012, 09:21 AMPlatoRe: It's limit.
- Aug 19th 2012, 03:08 PMTiffanyRe: It's limit.
Sorry my bad. It is -1 not -2 :)

- Aug 19th 2012, 03:30 PMPlatoRe: It's limit.
- Aug 19th 2012, 04:37 PMDevenoRe: It's limit.
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. - Aug 21st 2012, 12:29 AMVlasevRe: It's limit.
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.