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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.

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    $\begingroup$ I guess you meant when $ h \rightarrow 0$ $\endgroup$ – Atmos Feb 3 '18 at 11:35
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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.

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

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  • $\begingroup$ Hi, thanks! In the numerator we have $f(x)$ and $f(a)$. Let $x=a+h$ so $f(x)=f(a+h)$ and for $f(a)$ I have $x=a+h\iff a=x-h$ so $f(a)=f(x-h)$. And finally I have $\frac{f(a+h)-f(x-h)}{h}$. Why didn't you change the variable $a$ in $f(a)$ with $a=x-h$? $\endgroup$ – JDoeDoe Feb 3 '18 at 13:14
  • $\begingroup$ $a$ is a constant, to show the result you want to show, just let it like it is. $x-h=a$ so it is the same as you wrote but does not correspond to the definition you want to have $\endgroup$ – Atmos Feb 3 '18 at 13:31
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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 $x$?

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  • $\begingroup$ Hi, thanks! I started with $h=x-a$ and in the numerator I have $f(x)$ and $f(a)$. So I solved for $x$ in $h=x-a $ so $f(x)=f(a+h)$ and then I solved for $a$ in $h=x-a$ so $f(a)=f(x-h)$. Therefore I got $\frac{f(a+h)-f(x-h)}{h}$. Isn't the change of variable $a=x-h$ in $f(a)$ correct? $\endgroup$ – JDoeDoe Feb 3 '18 at 13:19
  • $\begingroup$ $a$ is a constant. So if you change equivalent like that, it should be $\lim_{h\rightarrow 0, x-h=a}\frac{f(a+h)-f(x-h)}{h}$. You got $\lim_{h\rightarrow 0}\frac{f(a+h)-f(x-h)}{h}$ missing the equation $x-h=a$($a$ is a constant). $\endgroup$ – liaoyulei Feb 6 '18 at 4:40

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