Question about properties of limits when changing variables Given that $\lim_{h \to 0} u = 0$, is it true and possible to prove that $\lim_{h \to 0} \frac{f(x+u) - f(x)}{u} = f'(x)$?
I understand that it is fairly intuitive, but what would be the rigorous proof for $\lim_{h \to 0} \frac{f(x+u)-f(x)}{u} = \lim_{u \to 0} \frac{f(x+u) 
- f(x)}{u}$
I came across this problem, when I was trying to rigorously prove the chain rule. I started with $\frac{d}{dx}[f(g(x))] = \lim_{h \to 0} \frac {f(g(x+h)) - f(g(x)}{h}$, then I let $\epsilon = g(x+h) - g(x)$, and substituted. $\lim_{h \to 0} \frac {f(g(x)+\epsilon)-f(x)}{h} = \lim_{h \to 0} \frac{\epsilon}{\epsilon} \cdot \frac{(g(x)+\epsilon) -f(x)}{h} = \lim_{h \to 0}\frac{\epsilon}{h} \cdot \lim_{h \to 0}\frac{f(g(x)+\epsilon)-f(g(x)}{\epsilon} = g'(x) \cdot \lim_{h \to 0}\frac{f(g(x)+\epsilon)-f(g(x)}{\epsilon} $ Here, I really want to just say that that final limit equals $f'(g(x))$, however,I am not sure whether I am allowed to do that.
 A: Here's what you can show:

Let $x\in\Bbb{R}$ be a given point, $f,u:\Bbb{R}\to\Bbb{R}$ be given functions such that $f$ is differentiable at the point $x$, and $\lim\limits_{h\to 0}u(h)=0$, and such that there is some $\delta>0$ such that for all $0<|h|<\delta$, we have that $u(h)\neq 0$. Then, we have that
\begin{align}
\lim_{h\to 0}\frac{f(x+u(h))-f(x)}{u(h)}=f'(x).
\end{align}

Actually, you don't need the domains of $f,u$ to be all of $\Bbb{R}$, I just stated it like this for convenience so that all the compositions are defined.
To prove the statement, define two new functions $F:\Bbb{R}\to\Bbb{R}$, $U:(-\delta,\delta)\to\Bbb{R}$,
\begin{align}
F(y)&=
\begin{cases}
\frac{f(x+y)-f(x)}{y}&\text{if $y\neq 0$}\\
f'(x)&\text{if $y=0$}
\end{cases}
\quad\text{and}\quad
U(h)=
\begin{cases}
u(h)&\text{if $h\neq 0$}\\
0&\text{if $h=0$}
\end{cases}
\end{align}
Now, we have $\lim\limits_{y\to 0}F(y)=\lim\limits_{y\to 0}\frac{f(x+y)-f(x)}{y}=f'(x)=F(0)$, because $f$ is differentiable at the point $x$. This says $F$ is continuous at $0$. Next, we have $\lim\limits_{h\to 0}U(h)=\lim\limits_{h\to 0}u(h)=0=U(0)$, which says that $U$ is continuous at $0$, and $U(0)=0$.
Therefore, we can now use the following fact about compositions of continuous functions: $U$ is continuous at $0$, and $F$ is continuous at $U(0)=0$, therefore, $F\circ U$ is continuous at $0$. Now what is the function $F\circ U$? It is simply
\begin{align}
(F\circ U)(h)&=F(U(h))=\begin{cases}
\frac{f(x+u(h))-f(x)}{u(h)}&\text{if $h\neq 0$}\\
f'(x)&\text{if $h=0$}
\end{cases}
\end{align}
So,
\begin{align}
\lim_{h\to 0}\frac{f(x+u(h))-f(x)}{u(h)}=\lim_{h\to 0}(F\circ U)(h)=(F\circ U)(0)=f'(x).
\end{align}
This completes the proof.

The above proof may look complicated, but the only real ideas used are the following:

*

*If two functions are equal everywhere except at a point then their limits are equal. More precisely, suppose $f,g$ are two functions such that there is a $\delta'>0$ such that for all $x$ with $0<|x-a|<\delta'$, we have $f(x)=g(x)$. Then, $\lim\limits_{x\to a}f(x)=\lim\limits_{x\to a}g(x)$.

*Composition of continuous functions are continuous (at the respective points), i.e if $f$ is continuous at $a$ and $g$ is continuous at $f(a)$, then $g\circ f$ is continuous at $a$.

The key theorem you were missing I think is this second one about continuity of compositions. Anyway for the chain rule, you have to be slightly more careful regarding division by $0$; for the details, take a look at this answer. The point is to define new functions which are continuous at the respective points, by "filling in" the removable discontinuity, and then you just take limits.
