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Prove that if $f$ is differentiable at $c$, then $f'(c) = \lim_{h\to 0}{f(c+h)-f(c)\over h}$.

I did this: If f is differentiable at $c$, then it's continuous at $c$, which implies,

$\lim_{x\to c}$ $f(x)=f(c)$ if only if $\lim_{h\to 0}$ $f(c+h)-f(c)=0$.

Then dividing by $h$,

$$\lim_{h\to 0}{f(c+h)-f(c)\over h}=f'(c).$$

I don't know if this is correct, I feel it's kind of arbitrary.

Edit: The definition of derivative is as follows- If $f$ is differentiable at a point a then the following limit exists: $\lim_{x\to a} {f(x)-f(a)\over x-a}$.

cqfd
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MarioM21
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    How can you "prove" a definition? – stoic-santiago Oct 11 '20 at 03:55
  • I don't really know, this is in my homework from Calculus 2 course. – MarioM21 Oct 11 '20 at 03:59
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    I see. Is this the complete question? Is there some background to it that we should know? (probably using one of the consequences of this definition to prove it) – stoic-santiago Oct 11 '20 at 04:00
  • Yes it's the complete question, the background I think is that we know everything from real number axioms to limits and continuity, also we know that if a function is differentiable at an arbitrary point a, then it's continuous in a. (Also most of the calculus courses at my university are mostly based on Spivak). – MarioM21 Oct 11 '20 at 04:05
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    @MarioM21 Can you add in your question the definition that you're using in classes about differentiablity? –  Oct 11 '20 at 04:13
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    Yes, post the definition. And also post the question exactly word for word as it is written. – Ameet Sharma Oct 11 '20 at 04:15
  • The question it's exactly the two first lines. – MarioM21 Oct 11 '20 at 04:23
  • Question is exactly in first line. – zkutch Oct 11 '20 at 05:24

3 Answers3

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So, from what I understand, the question boils down to showing the equivalence of two definitions of the derivative. Here's what you begin with:

If $f$ is differentiable at $x=a$, the following limit exists and is equal to $f'(a)$: $\lim_{x\to a} {f(x)-f(a)\over x-a}$

Now, put $x-a=h$, that is $x = a + h$. $x \to a$ is the same as $h \to 0$, so you finally get:

$f'(a) = \lim_{h\to0} {f(a+h)-f(a)\over h}$ as desired.

stoic-santiago
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The definition of $f'(c)$ is that if $\lim_{h\to 0}\frac {f(c+h)-f(c)}{h}$ exists then this limit is $f'(c),$ and if this limit doesn't exist then $f'(c)$ doesn't exist.

"$f $ is differentiable at $c$" is just another way to say that $f'(c)$ exists.

DanielWainfleet
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Nevertheless, that answer is done and accepted, but looking comments under OP, let me say, that here we have question about equivalence of differentiability and existence of derivative:

  1. Differentiability (Rudin W. - Principles of mathematical analysis-(1976) 212-213 p.). Suppose $f$ is defined on $(a,b) \subset \mathbb{R}$. We say, that $f$ is differentiable in $x \in (a,b)$ if exists linear mapping $A:\mathbb{R} \to \mathbb{R}$, such that $$\lim\limits_{h \to 0}\frac{f(x+h)-f(x)-Ah}{h}=0$$ this linear mapping we call differential and denote $df(x)(h)=Ah$

  2. Derivative.(Same book as above 211p) We say, that $f$ have derivative in $x \in (a,b)$, when exists $$\lim\limits_{h \to 0}\frac{f(x+h)-f(x)}{h}=B$$ and $B$ we denote as $B=f'(x)$

Now first sentence of OP mean, that this 2 definitions are equivalent: if exists $A$ in first, then exists $B$ from second and they are equal. Reverse also.

So $df(x)(h)=f'(x)h$. More here.

zkutch
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    Constructive critique is welcome, especially when this one is unique correct answer. – zkutch Oct 11 '20 at 11:02
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    I didn't downvote, but I suspect the downvote is due to the fact that you never clarified in your question (or even in your comments) what definition of derivative you're using. What you asked to be proven is the definition of the derivative in virtually every elementary calculus text and in the vast majority of undergraduate level real analysis and advanced calculus texts. The linear approximation definition you're using here is typically used in texts that primarily deal with multidimensional differentiation or Banach space differentiation, which is not suggested by your question's context. – Dave L. Renfro Oct 11 '20 at 11:11
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    @Dave L. Renfro. Thanks for for answer, but let say, that if you are not who down vote, then may be he/she have another reason, which is quite interesting. As to your view, then book brought by me is classical math. analysis book for tens of years. Same definitions is classic also in Munkres, Zorich, Fichtenholz - I brought link above. On other hand, especially in last years, there may be appeared books where is not definition of differentiability and only derivative. These may give leak of knowledge for new generations of mathematicians. (Continued..) – zkutch Oct 11 '20 at 11:25
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    (Continuing..) As to your remark about context, then let me not agree, as I can bring large list of textbooks, where both definitions and their equivalence are brought for exactly one real variable. As to "you never clarified in your question what definition of derivative you're using", then it is just before one I brought. Nevertheless, I am grateful for comment and add mention of definition of derivative to answer. – zkutch Oct 11 '20 at 11:32
  • Thanks a lot, this is amazing! Such an abstract and deep idea, for what I was thinking was simply a very simple or absurd question. I thought of differentiability and derivative like if they were synonimous. – MarioM21 Oct 12 '20 at 06:02
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    @MarioM21. Your comment is the best possible score for my answer and the biggest reward for me. Ask more if/when you needed it. – zkutch Oct 12 '20 at 06:11