What Distinguishes the Existence of lim_{x \rightarrow x_0} f'(x) from f'(x_0)?

kittybobo1
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Homework Statement



I had a quick question concerning some definitions. What is the difference between lim_{x \rightarrow x_0} f'(x) to exist and for f'(x_0) to exist, definition wise?

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The Attempt at a Solution

 
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If the function is continuous then the limit as x approaches y of f is equal to f evaluated at y. Continunity is not a required property of derivatives (there are examples to show this).
 
To take a very simple example, let f(x)= x2 if x is not 1, f(1)= 2. For any x other than 1, f(x)= x2 in some interval around 1 and so it's derivative is 2x. limit as x goes to 1 of f'(x) is 2. But since f(x) is not continuous at x= 1, it is not differentiable there.
 
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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