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

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?

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