How to get find the limit approaching 0

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((Sqr(1+h)-1)/(h)

I got 1/2 as the solution.

Did I do it right?
 
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Well, you have the right answer.
 
Yes, it's correct. How did you get the answer though?
 
I always preferred the method of looking at my neighbors paper!
 
HallsofIvy said:
I always preferred the method of looking at my neighbors paper!

I like proof by intimidation more.
 
Or proof by inaccessible literature :biggrin:

For example, a proof can be found in the proceedings of the 4th Hungarian conference of 1912, by L. Krueger.
 

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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