Q:

What is the following limit $\lim_{x\to x_0,\;x \,< \,x_0}\frac{f(x)-f(x_0)}{x-x_0}$, given $f(x)$, $x_0$

$x_0=1$ $$\lim_{x\to x_0,\;x \,< \,x_0}\frac{f(x)-f(x_0)}{x-x_0}$$ where $f:R\to R,\;$ and $$f(x)=\left | x-1 \right |+\left | x+1 \right |$$
I calculate it and I get $\lim_{x\to 1,\;x \,< \,1} \frac{2-2}{x-1}$ which is $\frac{0}{0}$. I do not know what to say after this. Maybe I did something wrong but I thought that maybe that $\frac{0}{0}=0$ cause you have 0 divided by something almost $0$.

A:

This is good.
$\lim_\limits {x\to 1^-} \frac {2-2}{x-1}$ 
However this does not mean that $\frac 00 = 0$  It most certainly does not.  $\frac 00$ is "indeterminate"  It can really be anything.
However you can say: $\lim_\limits {x\to 1^-} \frac {0}{x-1} = 0$
When $x$ is in the neighborhood of $1$ (and less than $1$) the fraction equals 0.

