Limit of Indeterminate Form (0/0) using L'Hopital Rule

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Homework Statement


http://puu.sh/1irk2


Homework Equations





The Attempt at a Solution


I am having trouble doing this question. I tried doing the L'Hopital but didn't work. When I subbed in 0 in the function, I got 0/0.

first time:
I got that: http://puu.sh/1iroz

It still gave me the indeterminate form (0/0). Could someone give me a hint on solving this limit, I know the answer is 0 (used wolframalpha) but I am unsure why it is.

Thanks
 
Last edited by a moderator:
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Use L'Hopital's rulke again on the new limit.
 
planauts said:

Homework Statement


http://puu.sh/1irk2


Homework Equations





The Attempt at a Solution


I am having trouble doing this question. I tried doing the L'Hopital but didn't work. When I subbed in 0 in the function, I got 0/0.

first time:
I got that: http://puu.sh/1iroz

It still gave me the indeterminate form (0/0). Could someone give me a hint on solving this limit, I know the answer is 0 (used wolframalpha) but I am unsure why it is.

Thanks

Use algebra to simplify your answer. I don't think the limit is zero.
 
Last edited by a moderator:
I also did not get zero

I would not suggest using L'hospital's rule again, it can be simplified to give you a limit which is not zero I believe
 
Nevermind, you are right. It is -1. I redid it, when I simplified it, all the x's that led to the evil division of zero disappeared.

Thanks
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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