What is the Definition of Derivative and How is it Proved?

[fishturtle1]

Homework Statement

Suppose $f$ is differentiable at $a$. Prove $\lim_{h\rightarrow 0} \frac{f(a+h) - f(a)}{h} = f'(a)$

Relevant Equations

$f$ is differentiable at $a$ means $\lim_{x\rightarrow a} \frac{f(x) - f(a)}{x-a}$ exists and is finite.

$\lim_{x\rightarrow a} g(x) = L$ means for any sequence $(x_n)$ that converges to $a$, we have $\lim_{n\rightarrow\infty} g(x_n) = L$.

Alternatively, $\lim_{x\rightarrow a} g(x) = L$ means for all $\varepsilon > 0$ there exists $\delta = \delta(\varepsilon) > 0$ such that $\vert x - a \vert < \delta$ implies $\vert g(x) - L \vert < \varepsilon$.

Proof: By definition of derivative,
$$f'(a) = \lim_{x\rightarrow a}\frac{f(x) - f(a)}{x - a}$$
exists and is finite. Let $(x_n)$ be any sequence that converges to $a$. By definition of limit, we have $$\lim_{x_n\rightarrow a} \frac{f(x_n) - f(a)}{x_n - a} = f'(a)$$. By definition of convergence, for all $\varepsilon > 0$ there exists $N = N(\varepsilon) > 0$ such that $n > N$ implies $\vert x_n - a \vert < \varepsilon$, i.e. $\lim_{n\rightarrow\infty} (x_n - a) = 0$. Let $h = x_n - a$? Then
$$\lim_{h \rightarrow 0} \frac{f(a + h) - f(a)}{h} = f'(a)$$. []

I'm pretty sure this is wrong because I said $h = x_n - a$, but $h$ is a constant and $x_n - a$ is a sequence.. also I don't think I can substitute $h$ under the limit like I did, but I'm not sure.

 

[member 587159]

You are mixing two characterisations of limits. I think it is better to work with the ordinary $\epsilon-\delta$-definition of limit in this case, and not with the sequential characterisation. But the right idea can definitely be observed in your proof, although as written it contains multiple mistakes. Here is how I would write it:

Let $\epsilon > 0$. Since $f'(a)$ exists, there is $\delta > 0$ such that

$$\forall x \in\mathbb{R}: \left(0 < |x-a| < \delta \implies \left|\frac{f(x)-f(a)}{x-a}-f'(a)\right| < \epsilon\right) \quad (*)$$

Let $h \in \mathbb{R}$ with $0< |h|= |h-0| <\delta$. Then

$$\left|\frac{f(a+h)-f(a)}{h}-f'(a)\right| < \epsilon$$

since $|(a+h)-a| = |h| < \delta$ (i.e. we apply $(*)$ with $x= a+h$).

Thus, we have proven that for each $\epsilon >0$, there is $\delta > 0$ such that for all $h \in \mathbb{R}$ with $0 < |h| < \delta$, we have $\left|\frac{f(a+h)-f(a)}{h}-f'(a)\right|< \epsilon$.

This is precisely the definition of $\lim_{h \to 0} \frac{f(a+h)-f(a)}{h} = f'(a)$ and we are done!
____________

Similar exercise, to see if you understood this one:

Show that for a function $f: \mathbb{R} \to \mathbb{R}$ and a real number $x \in \mathbb{R}$:

$$\lim_{h \to 0} f(x+h) \mathrm{\ exists \ and \ equals \ } f(x) \iff \lim_{y \to x} f(y) \mathrm{\ exists \ and \ equals \ } f(x)$$

 

[fishturtle1]

You are mixing two characterisations of limits. I think it is better to work with the ordinary $\epsilon-\delta$-definition of limit in this case, and not with the sequential characterisation. But the right idea can definitely be observed in your proof, although as written it contains multiple mistakes. Here is how I would write it:

Let $\epsilon > 0$. Since $f'(a)$ exists, there is $\delta > 0$ such that

$$\forall x \in\mathbb{R}: \left(0 < |x-a| < \delta \implies \left|\frac{f(x)-f(a)}{x-a}-f'(a)\right| < \epsilon\right) \quad (*)$$

Let $h \in \mathbb{R}$ with $0< |h|= |h-0| <\delta$. Then

$$\left|\frac{f(a+h)-f(a)}{h}-f'(a)\right| < \epsilon$$

since $|(a+h)-a| = |h| < \delta$ (i.e. we apply $(*)$ with $x= a+h$).

Thus, we have proven that for each $\epsilon >0$, there is $\delta > 0$ such that for all $h \in \mathbb{R}$ with $0 < |h| < \delta$, we have $\left|\frac{f(a+h)-f(a)}{h}-f'(a)\right|< \epsilon$.

This is precisely the definition of $\lim_{h \to 0} \frac{f(a+h)-f(a)}{h} = f'(a)$ and we are done!
____________

Similar exercise, to see if you understood this one:

Show that for a function $f: \mathbb{R} \to \mathbb{R}$ and a real number $x \in \mathbb{R}$:

$$\lim_{h \to 0} f(x+h) \mathrm{\ exists \ and \ equals \ } f(x) \iff \lim_{y \to x} f(y) \mathrm{\ exists \ and \ equals \ } f(x)$$

Thank you so much, this clears up my confusion.

Edit: just read the edit, will try.

 

[fishturtle1]

Show that for a function $f : \mathbb{R} \rightarrow \mathbb{R}$ and a real number $x \in \mathbb{R}$:
$\lim_{h\rightarrow 0}f(x+h)$ exists and equals $f(x) \iff \lim_{y\rightarrow x}f(y)$ exists and equals $f(x)$.

Proof: $(\Rightarrow)$ Let $\varepsilon > 0$. Then there is $\delta = \delta(\varepsilon) > 0$ such that $$\vert h - 0 \vert < \delta$$ implies $$\vert f(x+h) - f(x) \vert < \varepsilon$$ Then, for all $y \in \mathbb{R}$, if $$\vert y - x \vert = \vert h \vert < \delta$$ we have $$\vert f(x + (y - x)) - f(x) \vert = \vert f(y) - f(x) \vert < \varepsilon$$ This shows that $\lim_{y \rightarrow x} f(y) = f(x)$.

$(\Leftarrow)$ Let $\varepsilon > 0$. Then there is $\delta = \delta(\varepsilon) > 0$ such that for any $y \in \mathbb{R}$, if $$\vert y - x \vert < \delta$$ then $$\vert f(y) - f(x) \vert < \varepsilon$$
So, for any $h \in \mathbb{R}$, if
$$\vert h - 0 \vert = \vert y - x \vert < \delta$$
then
$$\vert f(x + h) - f(x) \vert < \varepsilon$$
This shows $\lim_{h\rightarrow 0} f(x + h) = f(x)$. []

 

[member 587159]

Proof: $(\Rightarrow)$ Let $\varepsilon > 0$. Then there is $\delta = \delta(\varepsilon) > 0$ such that $$\vert h - 0 \vert < \delta$$ implies $$\vert f(x+h) - f(x) \vert < \varepsilon$$ Then, for all $y \in \mathbb{R}$, if $$\vert y - x \vert = \vert h \vert < \delta$$ we have $$\vert f(x + (y - x)) - f(x) \vert = \vert f(y) - f(x) \vert < \varepsilon$$ This shows that $\lim_{y \rightarrow x} f(y) = f(x)$.

$(\Leftarrow)$ Let $\varepsilon > 0$. Then there is $\delta = \delta(\varepsilon) > 0$ such that for any $y \in \mathbb{R}$, if $$\vert y - x \vert < \delta$$ then $$\vert f(y) - f(x) \vert < \varepsilon$$
So, for any $h \in \mathbb{R}$, if
$$\vert h - 0 \vert = \vert y - x \vert < \delta$$
then
$$\vert f(x + h) - f(x) \vert < \varepsilon$$
This shows $\lim_{h\rightarrow 0} f(x + h) = f(x)$. []

Correct! But I would change the part"$|y-x|=|h|<\delta$" simply to "$|y-x|<\delta$" because you didn't define $h$ anywhere. You can just add a line with "we apply the previous inequality with $h=y-x$" to make it clear what you are doing. But this is just a small nitpick. Everybody will understand what you are doing and that'the most important thing.

 

[fishturtle1]

Correct! But I would change the part"$|y-x|=|h|<\delta$" simply to "$|y-x|<\delta$" because you didn't define $h$ anywhere. You can just add a line with "we apply the previous inequality with $h=y-x$" to make it clear what you are doing. But this is just a small nitpick. Everybody will understand what you are doing and that'the most important thing.

OK, I see what you mean. Thank you again for your help.

 

[epenguin]

Is it just me, am I the only one missing the point of all this?

The student states what differentiable means, states the definition of f', to prove the required formula is a trivial substitution in the definition formula, the two formulae say the same thing really.

What am I missin?

 

[member 587159]

Is it just me, am I the only one missing the point of all this?

The student states what differentiable means, states the definition of f', to prove the required formula is a trivial substitution in the definition formula, the two formulae say the same thing really.

What am I missin?

Yes, the two formulae say the same thing formally. But one must prove formally that one is allowed to make the substitution. This is the whole point of the exercise.