Proving Existence of lim (x--->c) {f(x)} = l

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suppose that f : (a,b)\{c} ----> real numbers is a function such that

lim (x--->c+) {f(x)} and lim (x--->c-) {f(x)} both exist and are equal to a common value l.

how can we actually prove that lim (x--->c) {f(x)} exists and that it equals l?
 
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Try a \epsilon,\delta proof. It's not to hard if you do just that...
 
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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