# Limit problem

1. Jan 8, 2006

### apchemstudent

How do you find the limit to:

Lim
x->0+ sqrtx/(1-cosx)

I use the L'Hopital rule and i get

lim
x->0+ 1/(2sqrtx*sinx)

and I can't get anywhere after this. I even tried multiplying by sqrtx/sqrtx to get rid of the sqrtx at the beginning, but it still doesn't work. Please help, thanks.

2. Jan 8, 2006

### shmoe

What's sqrt(x) going to as x->0+? sin(x)?

3. Jan 9, 2006

### dextercioby

Use the asymptotic expansion of cosine

$$\cos x \simeq 1-\frac{x^{2}}{2}$$ when "x" is close to zero.

Daniel.

4. Jan 9, 2006

### fourier jr

it's still in the form infinity * infinity isn't it? got to use l'hopital's rule one more time. i think the answer is 0.

5. Jan 9, 2006

### Hurkyl

Staff Emeritus
Check your list of indeterminate forms.

6. Jan 14, 2006

### Ishu

it is simple I guess...use the
"when top is going to 0 and bottom is going to soem where else."......I forgot the rule...But the answer is +0

7. Jan 14, 2006

### arildno

Just rewrite your expression as
$$\frac{\sqrt{x}}{1-\cos(x)}=\frac{1}{x^{\frac{3}{2}}}(\frac{x}{\sin(x)})^{2}(1+\cos(x))$$
Then it is easy to see what you should get.
Alternatively, use dexter's approach..

8. Jan 14, 2006

### Ishu

No you dont have to make it hard..beleive me it so simple....we did in calc 1 I forgot the rule name.

9. Jan 14, 2006

### arildno

10. Jan 14, 2006

### Ishu

opps didnt noticed bottom was going to 0 too...haha
then you will have to use l'hopital rule.

11. Jan 14, 2006

### mathwonk

well i just graphed them and looked at the graphs. sqrt(x) is really big compared to x, while 1-cos(x) is really small, so I think the limit is infinity. but now i will do the math, and no doubt come back chagrined.

12. Jan 14, 2006

### mathwonk

no i seem to be right, i.e. one use of l'hopital gives it, yes?

13. Jan 14, 2006

### mathwonk

or just write it as (sqrt(x)/x). (x/[1-cos(x))]) and both factors go to + infinity, the first trivially and the second by l'hopital.