Two definitions for derivative

Question

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

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 $x = c+h$?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

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 there is a point $q\in Y$ with the following property: For every $\epsilon > 0$ then there exists a $\delta > 0$ such that $d_Y(f(x), q) < \epsilon$ for all points $x \in E$ for which $0 < d_X(x, p) < \delta$

But what that means is (fleablood'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 $d-\epsilon < f(x) < d+ \epsilon$

$\lim_{x\rightarrow c}\frac {f(x) -f(c)}{x-c} = f'(c)=K$ 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.

Answer 5

As the old saying goes,

don't ask what your *StackExchange* can do for you but what you can do for your *StackExchange*

I would like to formalize the proof under these 2 rationales from above

1. This is not a derivative problem. This is, instead, a problem of limit. ~fleablood
2. Sequential limit criterion clears this up immediately. ~Yves Daoust

Without further adieu, let's begin with some definitions:

1. $g:\mathbb{A}\to\mathbb{R}$
2. $c\in\mathbb{A}$
3. g is differentiable at c

$\begin{align*} \lim_{h\to 0}\frac{g(c+h)-g(c)}{h}&\implies \forall (h_n)\rightarrow 0\land h_n\neq 0,\lim_{n\to\infty}\frac{g(c+h_n)-g(c)}{h_n}\text{ exists}\\ &\implies \forall (a_n)\subseteq\mathbb{A}\land a_n\rightarrow c, a_n-c\rightarrow 0, a_n-c\neq 0\\ &\implies\lim_{n\to\infty}\frac{g(c+a_n-c)-g(c)}{a_n-c}\text{ exists}\\ &\implies\forall (a_n)\subseteq\mathbb{A}\land a_n\rightarrow c\land a_n\neq c, \lim_{n\to\infty}\frac{g(a_n)-g(c)}{a_n-c}\text{ exists}\\ &\implies\lim_{a\to c}\frac{g(a)-g(c)}{a-c}\text{ exists}\\ \lim_{a\to c}\frac{g(a)-g(c)}{a-c}&\implies\forall a_n\rightarrow c\land a_n\neq c,\lim_{n\to\infty}\frac{g(a_n)-g(c)}{a_n-c}\text{ exists}\\ &\implies\forall h_n\rightarrow 0\land h_n\neq 0,c+h_n\rightarrow\land c+h_n\neq c\\ &\implies\lim_{n\to\infty}\frac{g(c+h_n)-g(c)}{c+h_n-c}\text{ exists}\\ &\implies\lim_{h\to 0}\frac{g(c+h)-(c)}{h}\text{ exists} \end{align*}$

There you have it, a bidirectional proof. Please point out the mistakes if there are. Thanks.