proof of chain rule Is my proof correct?
show: $(g\circ f)'(x_0)=g'(y_0)f'(x_0)$
Since $f'(x_0) = \lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0}$
and Since $g'(y_0) = \lim_{y\to y_0} \frac{g(y)-g(y_0)}{y-y_0}$
Multiply them to get:
$$g'(y_0)f'(x_0) = \lim_{y\to y_0} \frac{g(y)-g(y_0)}{y-y_0} \lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0}$$
since $y=f(x)$ and $y_0 = f(x_0)$, we have: 
$$g'(f(x_0))f'(x_0) = \lim_{f(x)\to f(x_0)} \lim_{x\to x_0}  \frac{g(f(x))-g(f(x_0))}{f(x)-f(x_0)} \frac{f(x)-f(x_0)}{x-x_0}$$
$$g'(f(x_0))f'(x_0) = \lim_{f(x)\to f(x_0)} \lim_{x\to x_0}  \frac{g(f(x))-g(f(x_0))}{x-x_0}$$
$$g'(f(x_0))f'(x_0) = \lim_{f(x)\to f(x_0)} (g\circ f)'(x_0)$$
not sure how to get rid of that limit. 
proof: (2nd try)
To prove the chain rule, I use Newton's approximation.
We need to show that:
$$\forall \epsilon >0\text{   }\exists \delta \text{  such that  } |g(f(x))-g(f(x_0))-f'(x_0)g'(y_0)(x-x_0)-|\leq \epsilon |x-x_0| \text{ whenever }|x-x_0|\leq \delta\text{and} x\in X$$
Since $g(y)$ is differentiable at $y_0$, we Know:
$$\forall \epsilon_1>0 \text{   }\exists \delta_1 \text{such that} 
|g(y)-g(y_0)-g'(y_0)(y-y_0)|\leq\epsilon_{1}|y-y_0| \text{ whenever }|y-y_0|\leq \delta_1\text{and} y\in Y$$
for $x\neq x_0$
choose $\delta_1 =f'(x_0)(x-x_0)$ and $\epsilon_1 = \frac{\epsilon }{f'(x_0)(x-x_0)} $ 
Proof (3rd try):
$$RHS = lim_{y\to y_0,y\in Y-\{y_o\}}\frac{g(y)-g(y_0)}{y-y_0}lim_{x\to x_0,x\in X-\{x_o\}}\frac{f(x)-f(x_0)}{x-x_0}$$ 
$$RHS = lim_{y\to y_0,y\in Y-\{y_o\}}\frac{g(y)-g(y_0)}{y-y_0}lim_{x\to x_0,x\in X-\{x_o\}}\frac{y-y_0}{x-x_0}$$
$$RHS = lim_{y\to y_0,y\in Y-\{y_o\}}\frac{y-y_0}{y-y_0}lim_{x\to x_0,x\in X-\{x_o\}}\frac{g(f(x))-g(f(x_0))}{x-x_0}$$ 
$$RHS = lim_{x\to x_0,x\in X-\{x_o\}}\frac{g(f(x))-g(f(x_0))}{x-x_0}=LHS$$ 
Is this correct?
 A: No,  $\lim_{f(x)\to f(x_0)}$ doesn't make any sense and if $f$ is constant, you divide by $0$ in your proof.
One way of proving this is to set
$$w(x)=\begin{cases}\frac{g(f(x))-g(f(x_0))}{f(x)-f(x_0)}&\text{if }f(x)\ne f(x_0)\\g'(f(x_0))&\text{else}\end{cases}$$
Let's prove the continuity of $w(x)$ at $x_0$. We know $g'(f(x_0))$ exists, so for all $\varepsilon>0$ there's $\delta>0$ such that $|y-f(x_0)|<\delta\implies\left|\frac{g(f(x))-g(f(x_0))}{f(x)-f(x_0)}-g'(f(x_0))\right|<\varepsilon$ for any $y\ne f(x_0)$.
So for any sequence $x_n\to x_0$ from the continuity of $f(x)$ at $x_0$ there's some $N$ such that for all $n>N$ it's true that $|f(x_n)-f(x_0)|<\delta$. Now either $f(x)=f(x_0)$ and $w(x_n)=g'(f(x_0))$, or $f(x)\ne f(x_0)$ and $|w(x_n)-g'(f(x_0))|<\varepsilon$.
Now it's easy to make your proof rigorous:
\begin{align*}\lim_{x\to x_0}\frac{g(f(x))-g(f(x_0))}{x-x_0}&=\lim_{x\to x_0}w(x)\frac{f(x)-f(x_0)}{x-x_0}=\lim_{x\to x_0}w(x)\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}\\&=g'(f(x_0))f'(x_0)\end{align*}
A: taking $\lim_{x \to x_0}$ implies $\lim_{y \to y_0}$ you don't need this extra limit to begin with
This would be like adding an extra degree of freedom to the problem. 
You can simply say
$g'(y_0)f'(x_0) = \lim_{x \to x_0}{[\frac{g(f(x)) - g(f(x_0))}{y - y_0} \frac{f(x) -f(x_0)}{x - x_0}]}$
A: Here's a neat well-written one I just found: http://kruel.co/math/chainrule.pdf
