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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- August 19th 2012, 09:59 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 . - August 19th 2012, 10:21 AMPlatoRe: It's limit.
- August 19th 2012, 04:08 PMTiffanyRe: It's limit.
Sorry my bad. It is -1 not -2 :)

- August 19th 2012, 04:30 PMPlatoRe: It's limit.
- August 19th 2012, 05:37 PMDevenoRe: It's limit.
perhaps you were a bit over-eager to use l'hopital?

we can use l'hopital once, to get:

but in the latter limit, the numerator goes to 1, while the denominator goes to 0, so the ratio is unbounded. - August 21st 2012, 01:29 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

for complex values . We'll show that is analytic. The condition with the Cauchy-Riemann equations is equivalent to saying that

for

If you calculate this, it is indeed equal to zero. So the function is analytic and we can take the limit as . The limit happens to be real and unbounded in the positive direction.