# Limit question with absolute value: $\lim_{x\to 4^{\large -}}\large \frac{x-4}{|x-4|}$

How would I solve the following limit, as $\,x\,$ approaches $\,4\,$ from the left? $$\lim_{x\to 4^{\large -}}\frac{x-4}{|x-4|}$$ Do I have to factor anything?

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As $x \to 4$ from the left, $\;x \lt 4 \implies x - 4 \lt 0 \implies |x-4|=(4-x) = -(x - 4).\;$

Substitute $\;-(x - 4)\;$ for $\;|x - 4|\;$ into the original limit, and divide.

$$\lim_{x\to 4^-}\frac{x-4}{|x-4|} = \; \lim_{x\to 4^-}\frac{(x-4)}{-(x - 4)} = -1.$$

after canceling the common factor in the numerator and denominator.

Similarly, as $x \to 4$ from the right, $x \gt 4 \implies (x - 4) > 0 \implies |x - 4| = (x - 4)$.

So $$\lim_{x\to 4^+}\frac{x-4}{|x-4|} = \; \lim_{x\to 4^+}\frac{(x-4)}{(x - 4)} = 1.$$

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thanks I understand now. – Fernando Martinez Jan 24 '13 at 21:44
You're welcome! – amWhy Jan 24 '13 at 21:44

Hint: If $x \lt 4, |x-4|=4-x$. Now you can just divide.

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$$\lim_{x\to 4^-}\frac{x-4}{|x-4|}=\lim_{x\to 4^-}\frac{x-4}{-(x-4)}=-1$$

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Yes, you do have to factor something. But you have to take a little bit of care.

If you want the left-hand limit, you have $x<4$. If $x<4$ then $x-4 < 0$ and so $|x-4|$ is actually the negative of $x-4$. In other words: $|x-4| \equiv -(x-4)$ for all $x<4$. Hence:

$$\frac{x-4}{|x-4|} \equiv \frac{x-4}{-(x-4)} \equiv -1$$

for all $x<4$. It follows that the limit is also $-1$. A similar argument shows that $|x-4|\equiv x-4$ for all $x>4$ and so the right-hand limit is $+1$.

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