1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
9
$\begingroup$

$7$ 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 $1$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.