# Chain rule proof help me dunno what's wrong

1. Mar 14, 2013

### Andrax

1. The problem statement, all variables and given/known data
this is kinda funny i've been strugglnig with the proof for 3 hours straight today
long story short(<--sorry couldn't find the best way t o say this ) i made it to this statement
lim(x->a) g(x) if this is equal to g(x) then my proof is done
lim(x->a) f(a) if this is equal to f(x) then my proof is done
what's on my mind : this is obviously NOT correct ,after 3 hours of pain i decided to check the txtbook found everything that ive doneto this part and they just write
lim(x->a) g(x)=g(x) and lim(x->a) f(a)=f(x) vthis made me pretty mad i have no idea how they've done this , i was good at limits i have no idea how is this correct if someone can help explain it to me please

forgot to metion i don't think this is important though they use h instead of x-a
chain rule = d/dx (f*g) = df/dxf*g+dg/dxg*f

2. Relevant equations
without reading the above why lim(x->a) g(x)=g(x)
and why lim(x->a) f(a)=f(x)

3. The attempt at a solution
what i've said up

Last edited: Mar 14, 2013
2. Mar 14, 2013

### Fredrik

Staff Emeritus
$\lim_{x\to a}g(x)$ is not equal to g(x). Note that this wouldn't even make sense, since x is a dummy variable in $\lim_{x\to a}g(x)$. This expression means the same thing as $\lim_{t\to a}g(t)$. It's equal to g(a) if and only if g is continuous at a. (Note that if g is differentiable at a, it is also continuous at a).

3. Mar 15, 2013

### Andrax

Oh i totally missed this part g is actually continous at a thanks