Why does $\lim_{h \to 0^-} \frac{f(x+h) - f(x)}{h} \neq \lim_{h \to 0} \frac{f(x-h) - f(x)}{h} $ I realize that the only reason one-sided limits arise is as a result of the $\epsilon-\delta$ definition of a limit, applied to the real field $\mathbb{R}$, and that one-sided limits aren't even well defined when you move to complex valued functions, but this is a simple question that I can't seem to answer :
Why is the following true (note the $0^{-}$ in the term on the left) :
$$\lim_{h \to 0^-} \frac{f(x+h) - f(x)}{h} \neq \lim_{h \to 0} \frac{f(x-h) - f(x)}{h} $$
Since $h \to 0$, for values of $h < 0$ (in this example), why does the the term $f(x+h)$, not become $f(x-h)$.
A simple example is given by $f(x) = |x|$
 $$|x| = \begin{cases}x & \text{if} & x\geq0 \\  -x & \text{if} & x<0 \end{cases}$$
$$f'_{-}(0) = \lim_{h \to 0^-} \frac{f(0-h) - f(x)}{h} = 1$$
Which is wrong, as the correct answer should be $-1$, so why is that we can't change the terms $f(x+h)$ to $f(x-h)$ as $h \to 0^{-}$ ?
 A: You’ve calculated the last limit incorrectly:
$$\lim_{h\to 0^-}\frac{f(0-h)-f(0)}h=-1\;,$$
not $1$. When $h$ is just a little less than $0$, $f(0-h)=f(-h)=|-h|=-h$, since $-h>0$, so the fraction is $\frac{-h}h=-1$.
However, it’s still not quite true that 
$$\lim_{h\to 0^-}\frac{f(x+h)-f(x)}h=\lim_{h\to 0}\frac{f(x-h)-f(x)}h\;:$$
it’s possible for the first limit to exist when the second does not. The correct statement is that
$$\lim_{h\to 0^-}\frac{f(x+h)-f(x)}h=\lim_{h\to 0^{\color{red}{+}}}\frac{f(x-h)-f(x)}{-h}\;,$$
so that both limits are one-sided.
A: When you are dealing with an expression like $\lim_{h \to 0}F(h)$ then you are dealing with values of $F(h)$ near $0$ (but not dealing with $F(0)$) and this means value of $F(h)$ for positive as well as negative $h$. When you deal with $\lim_{h \to 0^{-}}F(h)$ you are dealing with values of $F(h)$ near $0$ and $h$ is always negative.
In the expression $$\lim_{h \to 0^{-}}\frac{f(x + h) - f(x)}{h}\tag{1}$$ you are essentially dealing with values of $f$ at $x$ (i.e. $f(x)$) and values of $f$ at points which are near to but less that $x$ (i.e $f(x + h)$ and $x + h < x$). At the same time the denominator is negative.
Now check the expression $$\lim_{h \to 0}\frac{f(x - h) - f(x)}{h}\tag{2}$$ Here we are dealing with values of $f$ at $x$ as well as values of $f$ near $x$ and this time since $h$ can be positive as well as negative, we are dealing with values of $f$ at points less than $x$ as well as at points greater than $x$. Also if we are dealing with values of $f$ at points less than $x$ (so that $h$ is positive) then the denominator is positive and if we are dealing with values of $f$ at points greater than $x$ then denominator is negative.
You can clearly see that the bold statement in last to last paragraph does not match any of the two bold statements in last paragraph. And hence these expressions $(1)$ and $(2)$ are completely different things. They may be same by luck and not by any rule of logical inference.
The logically correct statement is $$\lim_{h \to 0^{-}}\frac{f(x + h) - f(x)}{h} = \lim_{h \to 0^{+}}\frac{f(x - h) - f(x)}{-h}$$ and you can check that they are referring to the same thing by using the kind of bold statements as I used in my earlier paragraphs. Or to be more technical/formal the expression on LHS gets transformed into that on RHS via the substitutions $h = -t$ and $h = t$ applied one after another in succession.
