Find the Limits of f(x)=(x^2-16)/(sqrt(x^2-8x+16)) at x=4

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Hi I was wondering if someone could help me with this limit problem:

Find the lim x->a-, lim x->a+ and lim x->a
for the f(x)=(x^2-16)/(squareroot(x^2-8x+16)) a=4

I've tried to simplify the problem but after that i don't know wut to do =(
This is wut I've done:
y=(x^2-16)/(squareroot(x-4)^2)
y=(x^2-16)/abs(x-4)

any help would be appreciated.
 
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y=(x^2-16)/abs(x-4)
= x+4,x>4
=-x-4,x<4
this is a piece wise function, mow you can easily find the left,right limit
Note:Limit of f(x) as x approaches 4 doesn't exist as left and right limit at x=4 are not equal

The more intuitive way is to draw the graph of the piecewise function!
 
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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