Finding the limit of an indeterminate using series, not lhopitals

[proace360]

Homework Statement

My calc teacher says that we should do the problem "by series", even though it is in the section of the book where the teach L'Hopital's rule.

For example, one of the questions is:

$$\lim_{x\rightarrow 0} \frac{sin(x)}{\sqrt{x}}$$

The Attempt at a Solution

Using L'Hopital's rule, you eventually get 0.

I managed to make a series for it...
$$\sum_{n=0}^\infty \frac{\frac{(-1)^n\*x^{2n+1}}{(2n+1)!}}{x^\frac{1}{2}}$$

but after that, i don't know what to do

 

[Office_Shredder]

You can turn that into

$$\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+.5}}{(2n+1)!}$$

by canceling the x in the denominator

 

[proace360]

yeah, i just simplified it on my homework, but after that, would you simply just plug 0 into x? it may give 0 for this problem, but for the other problems (after find the power series for them) it wouldn't give the right answer... so what would i do?

 

[HallsofIvy]

Another way to do this that does not use "L'Hopital" nor "series" is to write
$$\frac{sin(x)}{\sqrt{x}}= \sqrt{x}\frac{sin(x)}{x}$$
and use the fact that
[tex]\lim_{x\to 0}\frac{sin(x)}{x}= 1[/itex]

 

[proace360]

it would work that way too, but i just don't know what to do, as the professor specifically asked for it to be done "by series". and i am at a loss as to knowing how it can be done that way.

 

[proace360]

still need some help lol