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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 numerator only, one does the following:

$$g(x)={{\left|4-x\right|\over{x-4}} = \begin{cases}{4-x\over{x-4}} & x<4 \\ {-(4-x)\over{x-4}} & x>4 \end{cases}}$$

Eventually, of course, the pieces would be simplified, but I'll leave it like that for simplicity's sake.

Meanwhile, with an absolute value over an absolute value, I can't find the correct piecewise. When I take the limit of the function, I should get the answer $1$, but I can't do so without graphing the problem. Is there any way to create a correct piecewise version of this function?

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    $\begingroup$ hint:$4-x=-x+4=-(x-4)$ simplify $\endgroup$ Commented Jan 22, 2016 at 6:24

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You’re making it much too complicated. No matter what $x$ is, $4-x=-(x-4)$, so as long as $x\ne 4$,

$$\frac{4-x}{x-4}=-1\;,$$

and therefore

$$\left|\frac{4-x}{x-4}\right|=|-1|=1\;.$$

Of course the expression is undefined when $x=4$.

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  • $\begingroup$ I forgot to mention that I need to take the limit as x approaches 4, unfortunately. Thus, I figured the best way to "make the world safe for limits," as my teacher said, is to make it into a piecewise function. The problem on our worksheet looks like this: $\lim\limits_{x \to 4}{|4-x|\over{|x-4|}}$ $\endgroup$ Commented Jan 22, 2016 at 6:29
  • $\begingroup$ @Jenguinie: It doesn’t matter: the calculation above shows that the function is constantly $1$ for all $x\ne 4$, so each one-sided limit exists and is $1$. $\endgroup$ Commented Jan 22, 2016 at 6:30
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$f(x)=1$ for all $x\neq 4$:

$dom(f)=\{x\in\mathbb{R}:f(x)\mbox{ exists }\}=\{x\in\mathbb{R}:|x-4|\neq 0\}=\mathbb{R}\backslash\{4\}$

Now, if $x\in dom(f)$ (i.e. $x\neq 4$) we have $f(x)=\frac{|4-x|}{|x-4|}$. But all real $a$ we have $|a|=|-a|$. Hence $f(x)=\frac{|4-x|}{|x-4|}=\frac{|-(4-x)|}{|x-4|}=\frac{|x-4|}{|x-4|}=1$.

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