Find $f(7)$ if $f(x)=\frac{x-7}{|x-7|}$?

Question

I am trying to find $f(7)$ if $f(x)=\frac{x-7}{|x-7|}$. The problem I'm having, is that I don't know how to rewrite a function with an absolute value so that $f(7)$ exists.

I have tried multiplying both sides of the limit by conjugates, but it doesn't seem to get me anywhere.

$$\lim_{x \to 7^+} \frac{x-7}{|x-7|}=\lim_{x \to 7^+}\frac{x-7}{x-7}\cdot\frac{x+7}{x+7} = \lim_{x \to 7^+}\frac{x^2-49}{x^2-49}$$

$$\lim_{x \to 7^-} \frac{-(x-7)}{|-x+7|}=\lim_{x \to 7^-}\frac{-x+7}{x+7}\cdot\frac{x-7}{x-7}=\lim_{x \to 7^-}\frac{-x^2+14x-49}{x^2-49}$$

How can I solve this problem?

Answer 1

$f(x)=1$ if $x>7$ and $f(x)=-1$ if $x<7$ thus nor $f(7)$ neither $\lim_{x\to 7}f(x)$ does exist

Answer 2

$7$ isn't even in the domain of the function so it doesn't make sense to talk about $f(7)$. And you can't extend it via limits either, since the left limit is $-1$ (because $\frac{x-7}{-(x-7)} = -1$) and the right limit is $1$.