Two equivalent definitions of the derivative?

Question

I have seen two definitions of the derivative, but I don't grasp why why they are equivalent.

\begin{align} &\lim_{x\to a}\frac{f(x)-f(a)}{x-a} \tag 1\\ &\lim_{x\to 0}\frac{f(a+h)-f(a)}{h} \tag2 \end{align}

I tried a change of variable in $(1)$: $h=x-a$, so \begin{align} \frac{f(a+h) -f(x-h)}{h} \end{align} And the limit: For $x\to a$ we have $h=a-a=0$, so \begin{align} \lim_{h\to 0}\frac{f(a+h) -f(x-h)}{h} \end{align}

Is this correct so far? I' stuck here.

Answer 1

That's basically correct, just sub in $x-h=a$ in the last line:

$$ \begin{align} \lim_{h\to 0}\frac{f(a+h) -f(a)}{h} \end{align} $$

Make sure the substitution makes sense in terms of the formal definition of a limit as a sequence.

Answer 2

Your idea is good, if you put $x=a+h$ then when $x \rightarrow a$ correspond to $h \rightarrow 0$ and $$ \frac{f\left(x\right)-f\left(a\right)}{x-a}=\frac{f\left(a+h\right)-f\left(a\right)}{a+h-a}=\frac{f\left(a+h\right)-f\left(a\right)}{h} $$

Answer 3

The equivalent(2) should be: $$\lim_{h\rightarrow0}\frac{f(a+h)-f(a)}{h}$$ So change equivalent(1):$h=x-a$ so $x=a+h$, for $x\rightarrow a$ we have $h\rightarrow0$, so $$\lim_{h\rightarrow0}\frac{f(a+h)-f(a)}{h}$$

So why you get $f(x-h)$ here? What's the meaning of $x$?