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This may be a very stupid question and could be blatantly obvious, but I want to clear the confusion that I have about it. There are two equivalent definitions of the derivative:

Let $g: A \rightarrow \mathbf{R}$ be a function defined on an interval

$A$. Given $c \in A$, the derivative of $g$ at $c$ is defined by

$$g'(c) = \lim_{x \rightarrow c} \frac{g(x) – g(c)}{x-c}$$ provided

this limit exists.

and

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Let $g: A \rightarrow \mathbf{R}$ be a function defined on an interval

$A$. Given $c \in A$, the derivative of $g$ at $c$ is defined by

$$g'(c) = \lim_{h \rightarrow 0} \frac{g(c+h) – g(c)}{h}$$ provided

this limit exists.

My question is, what’s the formal reasoning why these two definitions are equivalent? For example, are we using the Algebraic Limit Theorem for functional limits? It is very clear to me “intuitively” why they are equivalently, i.e., simply let $x = c+h$ and one can see that in the first definition, as $x$ tends towards $c$ we “get” the expression $\frac{g(c)-g(c)}{c-c}$ while for the second definition, as $h$ tends towards $0$ we “get” the same expression $\frac{g(c)-g(c)}{c-c}$. But such reasoning is certainly not very rigorous and is very primitive, I wish to know why they are equivalent using formally justified reasons for each step in the process. For example, **why** can one substitute $x = c+h$ into the first definition and **why** after the substitution does the limiting variable change from $x$ to $h$?

EDIT: To be more precise, let $\phi(x)=\frac{g(x)-g(c)}{x-c}$ and let $\gamma(h) = \frac{g(c+h)-g(c)}{h}$, how can I prove that $\lim_{x \rightarrow c} \phi(x) = \lim_{h \rightarrow 0} \gamma(h)$ with the substitution $x = c+h$?

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The two functions $\phi$ and $\gamma$ you defined relates to each other by $\phi(x)=\gamma(x-c)$. So obviously, the following two limits exist simultaneously (or fail to exist simultaneously) and be equal that $$\lim\limits_{x\to c}\phi(x)=\lim\limits_{x\to c}\gamma(x-c).$$ The next problem is whether the following two limits exists simultaneously (or fail to exist simultaneously) and be equal: $$\lim\limits_{x\to c}\gamma(x-c),\lim\limits_{x\to 0}\gamma(x).$$ It turns out yes, because the expression $\gamma(x-c)$ is a composition of $\gamma$ and the map $x\mapsto x-c$. By this theorem, existence of $\lim\limits_{x\to 0}\gamma(x)$ implies existence of $\lim\limits_{x\to c}\gamma(x-c)$ and the limits are equal if they exist.

Similarly, you can prove that existence of $\lim\limits_{x\to c}\phi(x)$ implies existence of $\lim\limits_{x\to 0}\phi(x+c)$ and conclude the limits are equal if they exist (note that $\phi(x+c)=\gamma(x)$).

You need to check the the functions $x\mapsto x-c$ and $x\mapsto x+c$ satisfy Hypothesis $2$ as in the link, but it should be simple matter.

You can use the following fact (which would be the epsilon-delta result you are looking for):

Suppose that $\phi: A\to B$ and $\psi:B\to\mathbb{R}$ are two functions such that $\lim\limits_{x\to a} \phi(x)=b$ and $\lim\limits_{y\to b}\psi(y)=c.$

Then:

$$\lim\limits_{x\to a}(\psi\circ\phi)(x)=c.$$

Here, you can apply this result with

$$\psi(y)=\frac{g(y)-g(d)}{y-d} \text{ and } \phi(x)=d+x$$ in a first time, and with

$$\psi(y)=\frac{g(d+y)-g(d)}{y} \text{ and } \phi(x)=x-d$$ in a second time to show that the limits are the same and exist if and only if the other one exists.

In the $\delta,\epsilon$ definition of the limits, you work with neighborhoods.

Clearly

$$f(x),|x-c|<\delta$$ is strictly equivalent to

$$f(c+h),|h|<\delta$$ and you can perform the substitution $x\leftrightarrow c+h$ everywhere.

Okay, the question really isn’t about derivatives but about limits and why $\lim\limits_{a\rightarrow c} f(a) = \lim\limits_{c+h \rightarrow c}f(c+h) = \lim\limits_{h \rightarrow 0}f(c+h)$ are the same thing.

Well, as my second cousin twice removed never actually said “Let’s see what Mr. Rudin has to say”. (She frequently said “Let’s see what Mr. Webster has to say” or “Let’s see what Mrs. Rombauer has to say” but never “Mr. Rudin”.)

Rudin, Principals of Mathematical Analysis. Chapter 4 (Continuity) page 83.

Defin. Let X and Y be metric spaces: suppose $E \subset X$. $f$ maps

$E$ into $Y$ and $p$ is a limit point of $E$. We write

$\lim\limits_{x\rightarrow p} f(x) = q$ if ther is a point $q\in Y$

with the following property: For evvery $\epsilon > 0$ then there

exists a $\delta > 0$ such that $d_Y(f(x), q) < \epsilon$ for all

point $x \in E$ for which $0 < d_X(x, p) < \delta$

Okaaaay…..

That was a little more obtuse than I was expecting.

But what that means is (fleablod’s definition):

$\lim\limits_{x\rightarrow c}f(x) = d$ means for any small positive

value $\epsilon$ we can find a small positive value $\delta$, so that

whenever $c-\delta < x < c + \delta$ then $d-\epsilon < f(x) < d+ \epsilon$

So….

$\lim_{x\rightarrow c}\frac {f(x) -f(c)}{x-c} = f'(c)=K$ means that we have for any $\epsilon > 0$ we can find $\delta$ where

$c – \delta < x < c + \delta \implies K – \epsilon < \frac {f(x) -f(c)}{x-c} < K + \epsilon$

So let $c + h = x; h = x – c$. Then

$-\delta < h < \delta \implies$

$c – \delta < c + h < c+\delta \implies$

$c-\delta < x < c+ \delta \implies$

$K – \epsilon < \frac {f(x) – f(c)}{x- c} = \frac {f(c+h) – f(c)}{h} < K – \epsilon$.

So that would mean $\lim\limits_{h\rightarrow 0} \frac {f(c+h) – f(c)}{h} = K = f'(c)$.

In general:

$\lim\limits_{a\rightarrow c} f(a) = \lim\limits_{c+h \rightarrow c}f(c+h) = \lim\limits_{h \rightarrow 0}f(c+h)$

because ….

if $a = c + h$ then

$c – \delta < a < c + \delta \iff -\delta < h < \delta$.

So the definitions of $h\rightarrow 0$ and $c+h = a \rightarrow c$ are completely equivalent.

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