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How can we solve

$\lim_{h\rightarrow0}{\frac{|x+h|-|x|}{h}}$ with $x,h \in \mathbb{R}$, and similar limit problems with absolute values?

Note that although this represents $\frac{d}{dx}|x|$, I would like learn to solve this without any kind of differentiation/integration rules.

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Absolute values...of real or complex numbers, or norms of vectors? The notation you used is for norm, not for the usual absolute value... –  DonAntonio Jan 12 '13 at 15:34
    
Absolute values of real numbers. Please edit my post then because I thought the command for absolute value was \| \| –  xcrypt Jan 12 '13 at 15:35
    
Just use "|" without the backslash (or "\vert"). –  David Mitra Jan 12 '13 at 15:38
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2 Answers 2

up vote 3 down vote accepted

One approach to absolute values is to consider different cases.

If $x>0$, then for $h$ small enough, you will have $x + h > 0$, so $\lvert x\lvert = x$ and $\lvert x+h\lvert = x + h$, so $$ \lim_{h\to 0} \frac{\lvert x+h\lvert - \lvert x\lvert}{h} = \lim_{h\to 0} \frac{x+h - x}{h} = 1 $$

If $x<0$, then likewise $\lvert x\lvert = -x$ and $\lvert x+h\lvert = -(x+h)$.

If $x = 0$, you can consider the two limits where $h$ approaches $0$ from the left and from the right. For example: for $h$ approaching $0$ from the right, you have $h>0$, so $\lvert h\lvert = h$ and $$ \lim_{h\to 0^+} \frac{\lvert 0 + h\lvert - \lvert 0 \lvert}{h} = \lim_{h \to 0^+}\frac{h}{h} = 1. $$ Likewise you get $$ \lim_{h\to 0^-} \frac{\lvert 0 + h\lvert - \lvert 0 \lvert}{h} = \lim_{h \to 0^+}\frac{-h}{h} = -1. $$ What you have actually then shown is that the limit does not exist for $x=0$ since the right hand and the left hand limits to not equal. Hence $y = \lvert x\lvert$ is not differentiable at $x=0$.

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Hmm, this looks correct, but I was hoping for something that would steer me in the direction of |x|/x –  xcrypt Jan 12 '13 at 15:46
    
@xcrypt: Why do you want $\lvert x\lvert / x$? –  Thomas Jan 12 '13 at 15:47
    
Because then you generalized all cases with 1 equation/expression –  xcrypt Jan 12 '13 at 15:49
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@xcrypt: I am not sure that I understand. The absolute value is defined by cases, so it is natural that when the abs. values comes up in a proof, you might have to look at separate cases. But maybe someone else will come along and give a better argument. –  Thomas Jan 12 '13 at 15:51
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Besides Thomas's nice answer, you can use another way for finding what did you asked about. If $f(x)=|x|$ then $$f'^+(0)=\lim_{x\to 0^+}\frac{f(x)-f(0)}{x-0}=\lim_{x\to 0^+}\frac{|x|-|0|}{x}=\lim_{x\to 0^+}\frac{|x|}{x}=\lim_{x\to 0^+}\frac{+x}{x}=1$$ and $$f'^-(0)=\lim_{x\to 0^-}\frac{f(x)-f(0)}{x-0}=\lim_{x\to 0^-}\frac{|x|-|0|}{x}=\lim_{x\to 0^-}\frac{|x|}{x}=\lim_{x\to 0^-}\frac{-x}{x}=-1$$. This can show that $f(x)$ is not differentiable at $x=0$. I want to remark you in Thomas's first sentence. We can not write a general solving way for every absolute functions without knowing the function and the given point.

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+1 Using this definition of derivative things look much clearer. I'm also editing your answer and writing "Thomas" instead "tobias" –  DonAntonio Jan 12 '13 at 16:51
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@DonAntonio: Thanks Don. You always make me pleased. Sorry dear Thomas. Sorry for that. –  B. S. Jan 12 '13 at 17:19
    
@BabakSorouh: No problem :) –  Thomas Jan 12 '13 at 17:55
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