Finding the Limit of x->0+ sqrtx/(1-cosx)

[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.

 

[shmoe]

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

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

 

[dextercioby]

Use the asymptotic expansion of cosine

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

Daniel.

 

[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.

 

[Hurkyl]

Check your list of indeterminate forms.

 

[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

 

[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..

 

[Ishu]

No you don't have to make it hard..beleive me it so simple...we did in calc 1 I forgot the rule name.

 

[arildno]

So, why is your "answer" dead wrong, then?

 

[Ishu]

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

 

[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.

 

[mathwonk]

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

 

[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.