How to Solve Lim x->0+ sqrt(x)/(1-cos(x))?

[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.

 

[devious_]

\lim_{x \to 0+} \frac{1}{2 \sqrt{x} \sin x} = \infty

 

[apchemstudent]

\lim_{x \to 0+} \frac{1}{2 \sqrt{x} \sin x} = \infty

I take it as, there's no solution to this problem. Either I'm reading the question wrong, or the prof wrote the question wrong. This is because for any other problems in this exercise, the limit did exist, so I'm sceptical as to whether the answer is infinity.

 

[HallsofIvy]

L'hopital's rule works perfectly well. The second limit you got,
lim_{x\rightarrow 0^+} \frac{1}{2\sqrt{x}sin(x)}
does not exist (it goes to infinity) so the original limit,
lim_{x\rightarrow 0^+}\frac{\sqrt{x}}{1-cos(x)}
does not exist.

 

[Mindscrape]

Just plug in really small numbers (to the L'Hopital's), and you get 1/(really small) which would be really big, and as the numbers got smaller the answer would approach infinity.

 

[twoflower]

L'hopital's rule works perfectly well. The second limit you got,
lim_{x\rightarrow 0^+} \frac{1}{2\sqrt{x}sin(x)}
does not exist (it goes to infinity) so the original limit,
lim_{x\rightarrow 0^+}\frac{\sqrt{x}}{1-cos(x)}
does not exist.

Why doesn't it exist if it goes to infinity? Anyway, even if the l'Hospitalled limit didn't exist, I thought there would be nothing I could say about the original limit. The existence of limit

<br /> \frac{f&#039;(x)}{g&#039;(x)}<br />

is precondition for l'Hospital rule as far as we've been told in school.

 

[shmoe]

Why doesn't it exist if it goes to infinity?

Often (definitions do vary) saying a limit exists means it converges to a finite value. Saying a limit is infinity, or writing lim=+inf, is still saying the limit does not exist, but is giving more information on how it's not existing.

Anyway, even if the l'Hospitalled limit didn't exist, I thought there would be nothing I could say about the original limit.

If the limit is not existing in the "diverging to +infinity" (or -infinity) sense, then l'hospital's still gives information.

 

[twoflower]

Often (definitions do vary) saying a limit exists means it converges to a finite value. Saying a limit is infinity, or writing lim=+inf, is still saying the limit does not exist, but is giving more information on how it's not existing.

Yes, probably it's a matter of definition, for example we had defined it as follows:

<br /> \mbox{We say that }f\mbox{ has in point }a \in \mathbb{R}^*\mbox{ limit }A \in \mathbb{R}^*\mbox{ if...}<br />