Solving limit without using l'hopitals rule Find the limit as $h \to 0$ for $\frac{f(x)-f(x-h)}{h}$ at $x=0$, where $f(x) = x + |x|$ without using L'hopital rule.
Once I plug in the numbers and try to work it down I get brought to $\frac{-(-h + |-h|)}{h} = \frac{h-h}{h}$ which is $\frac{0}{h}$ and then if go to plug in $0$ for $h$ then I would get $0/0$ which is indeterminate. How should I approach this problem to get the answer?
 A: Two mistakes you are making:
one:  $|-h|\ne h$.  $|-h| =\begin{cases} h& h \ge 0\\-h& h < 0\end{cases}$.
two: And as $\frac 0h = 0$ for all $h \ne 0$, you have $\lim_{h\to 0} \frac 0h = \lim_{h\to 0} 0$.   The $0$ has already "canceled" the $h$ out so there is nothing to "plug in".
$\lim_{h\to 0^+} \frac {f(0) -f(0-h)}h = \lim_{h\to 0^+} \frac {-(-h + |h|)}h=\lim_{h\to 0^+} \frac {-(-h + h)}h =\lim_{h\to 0^+} \frac 0h =\lim_{h\to 0^+} 0 = 0$.
And $\lim_{h\to 0^-} \frac {f(0)-f(0-h)}h = \lim_{h\to 0^-} \frac {-(-h + |h|)}h=\lim_{h\to 0^-} \frac {-(-h - h)}h=\lim_{h\to 0^-} \frac {2h}h = \lim_{h\to 0^-} 2 = 2$.
(Note, you don't plug in $h = 0$ to get $\frac {2h}h=\frac 00$ because we aren't evaluating $\frac {2h}h$ when $h=0$.  We are taking the limit as $h$ is near but not equal to $0$.  So you don't plug in $h = 0$ for $\lim \frac 0h$ either.)
so the left hand limit is not equal to the right hand limit and the $\lim_{h\to 0} \frac {f(0) - f(0-h)}h$ does not actually exist.
A: In general,
$$
\lvert h\lvert=
\begin{cases}
h &,& h\geq0 \\
-h &,& h<0
\end{cases}
$$
If we want to find $\lim_{h\to 0}\lvert h\lvert$, we need to consider what happens as $h$ approaches $0$ from the right and the left.
If $h$ is approaching $0$ from the right, then $h>0$; that is, we are in the first case of $\lvert h\lvert$, and $$\lim_{h\to 0^+}\lvert h\lvert = \lim_{h\to 0^+}h.$$
If $h$ is approaching $0$ from the left, then $h<0$; that is, we are in the second case of $\lvert h\lvert$, and $$\lim_{h\to 0^-}\lvert h\lvert = \lim_{h\to 0^-}(-h).$$
Now, to your question. We want to find
$$\lim_{h\to 0}\frac{f(0)-f(0-h)}{h}=\lim_{h\to 0}\frac{h-\lvert h\lvert}{h},$$
after some simplifying.
For the right limit,
\begin{align*}
\lim_{h\to 0^+}\frac{h-\lvert h\lvert}{h}&=\lim_{h\to 0^+}\frac{h-h}{h}=\lim_{h\to 0^+}\frac{0h}{h}=\lim_{h\to 0^+}0=0
\end{align*}
For the left limit,
\begin{align*}
\lim_{h\to 0^-}\frac{h-\lvert h\lvert}{h}&=\lim_{h\to 0^-}\frac{h+h}{h}=\lim_{h\to 0^-}\frac{2h}{h}=\lim_{h\to 0^-}2=2
\end{align*}
Since the left and right limits don't agree, the limit does not exist.
