# Limit problem

## Main Question or Discussion Point

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.

shmoe
Homework Helper
apchemstudent said:
lim
x->0+ 1/(2sqrtx*sinx)
What's sqrt(x) going to as x->0+? sin(x)?

dextercioby
Homework Helper
Use the asymptotic expansion of cosine

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

Daniel.

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.

Hurkyl
Staff Emeritus
Gold Member
Check your list of indeterminate forms.

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

arildno
Homework Helper
Gold Member
Dearly Missed
$$\frac{\sqrt{x}}{1-\cos(x)}=\frac{1}{x^{\frac{3}{2}}}(\frac{x}{\sin(x)})^{2}(1+\cos(x))$$
Then it is easy to see what you should get.
Alternatively, use dexter's approach..

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

arildno
Homework Helper
Gold Member
Dearly Missed
So, why is your "answer" dead wrong, then? opps didnt noticed bottom was going to 0 too...haha
then you will have to use l'hopital rule.

mathwonk
Homework Helper
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.

mathwonk