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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

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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  • $\begingroup$ when you substitute $x \mapsto c+h$ you get $\frac{g(c+h) - g(c)}{c+h-c} = \frac{g(c+h)-g(c)}{h}$ and for the limit $c+h \to c \implies h \to 0$. They're the same. Also see here for a more rigorous explanation why we can use substitutions in limits. $\endgroup$
    – Dando18
    Jun 16 '17 at 15:04
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    $\begingroup$ As you seem to close to realizing, this is a "change of variable". In one case we have the variable $x$ tending to $c$, and in the other case we have variable $h$ tending to $0$. There are two aspects, given the substitution $x = c+h$. The easy one is to recognize the equality of the two forms of difference quotient under this substitution. The other aspect is only a little more sophisticated, that given the substitution, why is $x\to c$ equivalent to $h\to 0$? Which aspect is giving you difficulty? $\endgroup$
    – hardmath
    Jun 16 '17 at 15:07
  • $\begingroup$ Are you asking why addition works and why there is an additive identity? Also what is algebra limit theorem? Are you asking how variables work rigorously in math? Have you seen the epsilon delta definition of the limit? What are you asking exactly or what don't you understand? I could attempt to prove statements in terms of epsilon and delta, or prove that two statements have the same limit or maybe you know that and is a waist of everybody's time? $\endgroup$ Jun 16 '17 at 15:21
  • $\begingroup$ I can say this, they are equivalent in some high level aspect but the two are different mathematical statements. In logic there are tautology where one statement implies another, and that statement implies the first. A tautology is an equivalence relation. The derivative can be defined as a function taking a variable argument, a function, to some other set. So as a function it's graph defines it, these two definitions yield the same graph for inputs. So in that sense they define the same definition of derivative. $\endgroup$ Jun 16 '17 at 15:27
  • $\begingroup$ Would you be satisfied with a geometric argument? i.e. the derivative at $c$ is just the limit of the average rate of change of $g$ between two points as the distance between these points goes to zero. In the first case we name these points $c$ and $x$ and let $x \to c$. In the second case we name these points $c$ and $c+h$ and let $h \to 0$. We just changed the naming scheme. There is no reason the situations could be any different from each other. $\endgroup$
    – wgrenard
    Jun 16 '17 at 15:35
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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.

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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.

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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.

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    $\begingroup$ The "fact" you quoted is false. Consider $\phi$ defined by $\phi(x)=x\sin(\frac{1}{x})$ for $x\not=0$ and $\phi(0)=0$, and $\psi$ to be zero everywhere except at $0$. The composition function oscillates a lot as $x$ goes to $0$. $\endgroup$
    – edm
    Jun 16 '17 at 16:12
  • $\begingroup$ @edm: I think that your $\psi$ won't check the hypothesis I ask about its limit at $0$ ? $\endgroup$
    – Balloon
    Jun 16 '17 at 16:17
  • $\begingroup$ $\lim\limits_{x\to 0}\phi(x)=0$, $\lim\limits_{y\to 0}\psi(y)=0$, but $\lim\limits_{x\to 0}(\psi\circ\phi)(x)$ does not exist. $\endgroup$
    – edm
    Jun 16 '17 at 16:19
  • $\begingroup$ @edm: you are talking about pointed limits, here in my sense $\psi$ doesn't have any limit in $0$ since $\psi(\frac{1}{n})\to 0$ and $\psi(0)\to\text{"the constant of your choice different of $0$"}$ (edit: I don't know which is the english for "limite épointée", which is the french definition for taking the limit on $A\setminus \{a\}$, where $a$ is the point where you are interested taking the limit?) $\endgroup$
    – Balloon
    Jun 16 '17 at 16:25
  • $\begingroup$ Are you assuming that $\psi$ is continuous? From what I have learned, limits are defined as "pointed limits", and $\psi$ do not have to be continuous. $\endgroup$
    – edm
    Jun 16 '17 at 16:33
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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.

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It's kind of late, but I found that using the sequential limit criterion clears this up immediately because $(h_n) \to 0$ implies $(c + h_n) \to c$ and $(x_n) \to c$ implies $(x_n - c) \to 0$.

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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.

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  • $\begingroup$ Maybe you should "\implies" for logical implications, the difference between \rightarrow and \to is hard to see $\endgroup$
    – miracle173
    Jun 10 '21 at 4:28
  • $\begingroup$ I don't understand the meaning of a formula $\rightarrow \forall (h_n)\rightarrow 0\ldots$, can you explain? $\endgroup$
    – miracle173
    Jun 10 '21 at 4:33
  • $\begingroup$ @miracle173 I have changed all the rightarrows that basically indicates new line to implies. As to $\forall h_n\rightarrow 0$. I will pay you the complement by assuming you understand limits and universal quantifer. As such, I will go through the underlying presupposition--sequential criterion for limits. For all $ h_n\rightarrow 0\land \forall n\in\mathbb{N} h_n\neq 0\implies\lim_{n\to\infty}g(h_n)=c$ if and only if $\lim_{h\to 0}g(h)=c$. The idea behind this proof is that the existence of functional limit necessitates all sequential limits to converage at same point. $\endgroup$
    – zony_miu
    Jun 10 '21 at 7:48
  • $\begingroup$ @miracle173 I understand the confusion. $\forall h_n\rightarrow 0\land h_n\neq 0$ is absurd. I should have instead gone for this: For all sequences that converges to 0 and has no terms which are 0. This is my original intent. Bad use of quantifier. As to why each sequence should not contains 0, that's because limit is defined on a limit point. That limit point $c$ can be or not be in the domain. But it must be a limit point, which means $\forall\epsilon\in\mathbb{R^+}\exists a\in\mathbb{A},a\in V_{\epsilon}(c)\setminus\{c\}$. $\endgroup$
    – zony_miu
    Jun 10 '21 at 8:07

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