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Came across this question, for which I thought the limit should have been 16. Turns out it did not exist. I don't understand why: $$\lim_{x\to8}\frac{x^2-64}{|x-8|}$$

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    Hint: look at the one-sided limits. – 2017-02-10
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    Consider from lhs and rhs of 8 and see the mod – 2017-02-10
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    Because its positive 16 when approaching from above and negative when approaching from below. – 2017-02-10
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    Hint if $x\ne 8$ $(x^2-64)/(x-8) = x+8$ while $(x^2-64)/-(x-8)=-(x+8) $ – 2017-02-10
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    ohh I got it now. Thanks – 2017-02-10

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$$\lim_{x\to8}\frac{x^2-64}{|x-8|}=\lim_{x\to8}\frac{(x-8)(x+8)}{|x-8|}$$

If $x>8$ then $x-8>0$ and $|x-8|=x-8$ so $$\lim_{x\to8^{+}}\frac{x^2-64}{|x-8|}=\lim_{x\to8^{+}}\frac{(x-8)(x+8)}{x-8}=16$$ If $x<8$ then $x-8<0$ and $|x-8|=-(x-8)$ so $$\lim_{x\to8^{-}}\frac{x^2-64}{|x-8|}=\lim_{x\to8^{-}}\frac{(x-8)(x+8)}{-(x-8)}=-16$$

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    @John_Smith Thanks. – 2017-02-10
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    This part, although I understand what you mean, seems a little contradictory to me: |x - 8| = -(x - 8) ; isn't the stuff inside the parentheses evaluated first, then the absolute value operation applied? That should result in an always positive number the way I see it, 2hich is why I find that minus sign a little confusing – 2017-02-10
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    -(x-8) *is* a positive number if x <8. x-8 <0 so -(x-8) >0. And |x-8|=-(x-8). Or put another way x-8 < 0 so |x-8|= opposite value (x-8)=-(x-8). You aren't the first to find it confusing. – 2017-02-10
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When you see,

$$\frac{(x-8)}{|x-8|}$$

We cannot reduce this to $1$, if $x<8$ then we have $|x-8|=-(x-8)$ so then the fraction reduces to $-1$. I hope this helps you understand why the limit does not exist. From one direction the limit goes to $16(-1)$ and the other $16(1)$.

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I'll add to the other answers that have been given. Graph the function on the interval $(7, 9)$ using a graphing calculator or WolframAlpha.com. You'll see from the graph that that the limit does not exist.