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How would I come to the following conclusion analytically/algebraically:

One-sided limit

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The expression is constant near $2$ –  Hagen von Eitzen Sep 16 '13 at 22:29

2 Answers 2

up vote 2 down vote accepted

Since no one has answered the updated question, I guess I'll do it.

Since our limit is one-sided, we know that $x \lt 0$. This implies that $|x| = -x$, and so the expression inside the limit is $\frac{-x}{x} = -1$ if $x \neq 0$. Since for any $x \lt 0$ the expression is constantly $-1$, so is its limit.

Of course, the two sided limit doesn't exist, since the same argument shows that $\lim\limits_{x \to 0^+} \frac{|x|}{x} = 1$.

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When $x>2$, which happens when $x\rightarrow 2^+$ then $|x|=x$. Hence,

$\lim_{x\rightarrow 2^+}\frac{|x|}{x}=\lim_{x\rightarrow 2^+}\frac{x}{x}=\lim_{x\rightarrow 2^+}1=1$

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Sorry, I initially posted the wrong image. I have updated my question. –  Joel A. Christophel Sep 16 '13 at 22:34
Ok, but it is the same idea. From the left of $0$, $|x|=-x$. –  Daniel Montealegre Sep 16 '13 at 23:56

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