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I need to Find $\underset{x \to 2} \lim \frac{-4x+8}{|x-2|}$, but I'm not sure how to handle the absolute value in the denominator. If I can just multiply by $\frac{|x + 2|}{|x+2|}$, are there any special rules for that? How can I handle this?

I have tried:

$$\underset{x \to 2} \lim \frac{-4x+8}{|x-2|}$$

$$=\underset{x \to 2} \lim \frac{(-4x+8)(8+4x)}{|x-2|(8+4x)}$$

$$=\underset{x \to 2} \lim \frac{64-16x^2}{|x-2|(8+4x)}$$

But it's apparent I need to change $|x - 2|$ in some way, or I will always have a zero in the denominator.

I have also tried L'Hospital's Rule.

$$\underset{x \to 2} \lim \frac{-4x+8}{|x-2|}$$

$$=\underset{x \to 2} \lim \frac{\frac{d}{dx}-4x+8}{\frac{d}{dx}|x-2|}$$

but I learned $\frac{d}{dx}|x-2|=\frac{x-2}{|x-2|}$, which doesn't seem to help.

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  • $\begingroup$ What is the value when $x>2$? when $x<2$ ? $\endgroup$ – J. W. Tanner May 1 at 3:12
  • $\begingroup$ try to compute $\lim_{x\to 2^+}$ and $\lim_{x\to 2^-}$ $\endgroup$ – Zongxiang Yi May 1 at 3:13
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\begin{eqnarray} \lim_{x\to2}\frac{-4x+8}{|x-2|}&=&\lim_{x\to2}(-4)\cdot\frac{x-2}{|x-2|}\\\ &=&\lim_{x\to2}(-4)\text{sgn}(x-2)\\ \end{eqnarray}

Therefore

$$ \lim_{x\to2^+}(-4)\text{sgn}(x-2)=-4 $$

and

$$ \lim_{x\to2^-}(-4)\text{sgn}(x-2)=4 $$

So the limit does not exist at $2$.

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Note that if $$x\to 2^+\implies |x-2|=x-2$$ and if $$x\to 2^-\implies |x-2|=-(x-2)$$

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