# Tagged Questions

For questions about or involving the absolute value function.

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### Prove $||a| - |b||$ is less than or equal to $|a-b|$

I was given the hint to split it into two cases ($|a| - |b|$ being positive and negative) and then use the triangle inequality. However, since the triangle inequality says that $|a+b|$ is less than or ...
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### Prove that $|x|^2$ = $x^2$.

This is what I did, but I'm not sure if it's a good enough proof: Since $|x|$ is equal to $x$ when $x$ is greater than or equal to 0, and is equal to $-x$ when $x$ is less than 0, I said that $|x|^2$ ...
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### Absolute Value Rational Inequalities

Ok so I have the following two inequalities: $$\left| \frac{x+6}{x-2}\right| \leq 4$$ and \frac{x^2-1}{\left| x+2\right|} \leq 3(1-x) \end{...
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### How to prove that $\lim_{(x, y) \to (0, 0)} \left(\frac{1}{|x|} + \frac{1}{|y|}\right) = 0$? [closed]

Prove that $$\lim_{(x, y) \to (0, 0)} \left(\frac{1}{|x|} + \frac{1}{|y|}\right) = 0$$ I couldn't prove this. Please suggest a solution.
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### Prove $\lvert x\rvert$ = $\lvert-x\rvert$ for all real numbers $x$ [closed]

Been at this one for a long time. I'm trying to use the fact that $|x|$ = $x$ if $x$ is greater than or equal to 0, and $|x|$ = $-x$ if $x$ is less than 0. Then I want to split the proof into these 2 ...
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### Piecewise from Rational Absolute Value Function

How would one separate a function like the following into piecewise? $$f(x)={\left|4-x\right|\over{\left|x-4\right|}}$$ I've been taught that with a rational function with an absolute value in the ...
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### Proving $\max$ of $a, b$.

How do I prove that $$\max{\{a, b\}} = \frac{a + b + \left | a - b \right |}{2}$$ I have no idea how to even start the proof, any idea / intuition that can get me started is greatly appreciated. ...
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### Square divided by absolute value

First time posting on Math SE, with kind of a basic algebra question. Question Does the relation: $$\dfrac{(ab)^2}{|ab|} = \left|ab\right|$$ with $a,b \in \mathbb{R_{\ne 0}}$ always hold? It seems ...
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### Function with absolute value in denominator - limits

f(x)=(x-1)/(|2-x|-1) |2-x|= { |2-x|; x < +2} {-|2-x|; x >= +2} State domain, range and the equations of the asymptotes. D(f)= {x | x > 3 or x < 3} R(f)= {y | y > 1 or y <= -1} ...
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### Double absolute value inside integral

Any ideas as to go about doing this particular integral? $$\int\limits\limits_{-1}^{4}||x^2+x-6|-6| dx$$ I'm a bit confused as to how to consider the cases into account. My idea was to consider 4 ...
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### If $\lim_{x \to x_0} f(x) = L$, then $\lim_{x \to x_0} \lvert f(x)\rvert = \lvert L \rvert$.

If $\lim_{x \to x0} f(x) = L$, then $\lim_{x \to x0} \lvert f(x)\rvert = \lvert L \rvert$. I know this is true, because $\lvert f(x) \rvert - \lvert L \rvert <= \lvert f(x) - L \rvert < \epsilon$...
### Property of function $\varphi(x)=|x|$ on $\mathbb{R}$
Define $\varphi(x)=|x|$ on $[-1,1]$ and extend the definition of $\varphi(x)$ to all real $x$ by requiring that $\varphi(x+2)=\varphi(x).$ How do you prove that for any $s,t$  |\varphi(s)-\...