Limit Definition of Derivative as n Approaches Infinity

[songoku]

Homework Statement

Please see below

Relevant Equations

Limit
Sequence Theorem
Derivative from first principle

$f'(x_0)$ is defined as:
$$f'(x_0)=\lim_{h \rightarrow 0} \frac{f(x_0+h)-f(x_0)}{h}$$
or
$$f'(x_0)=\lim_{x \rightarrow x_0} \frac{f(x)-f(x_0)}{x-x_0}$$

I can imagine that as $n \rightarrow \infty$ the value of $f(b_n)$ and $f(a_n)$ will approach $f(x_0)$ so the value of the limit will be like tangent to graph $f(x)$ at point $x_0$

But I don't know how to do it mathematically. The definition I know for derivative is the limit approaches 0 while the question is n approaches infinity. How to relate the question to the definition?

Thanks

 

[anuttarasammyak]

How about
$$b_n-a_n:=h_n$$
$$h_n\rightarrow 0$$

 

[songoku]

How about
$$b_n-a_n:=h_n$$
$$h_n\rightarrow 0$$

I think I can solve the question using this hint.

Thank you very much anuttarasammyak

 

[PeroK]

Homework Statement:: Please see below
Relevant Equations:: Limit
Sequence Theorem
Derivative from first principle

View attachment 314745

$f'(x_0)$ is defined as:
$$f'(x_0)=\lim_{h \rightarrow 0} \frac{f(x_0+h)-f(x_0)}{h}$$
or
$$f'(x_0)=\lim_{x \rightarrow x_0} \frac{f(x)-f(x_0)}{x-x_0}$$

I can imagine that as $n \rightarrow \infty$ the value of $f(b_n)$ and $f(a_n)$ will approach $f(x_0)$ so the value of the limit will be like tangent to graph $f(x)$ at point $x_0$

But I don't know how to do it mathematically. The definition I know for derivative is the limit approaches 0 while the question is n approaches infinity. How to relate the question to the definition?

Thanks

Use $\epsilon-\delta$ to relate the two?

 

[songoku]

Use $\epsilon-\delta$ to relate the two?

Sorry I don't understand the hint.

I followed hint in post#2 and this is what I did:
Let $h_n=b_n -a_n$ and as $n \rightarrow \infty, h_n \rightarrow 0$
$$\lim_{n \rightarrow \infty} \frac{f(b_n)-f(a_n)}{b_n-a_n}$$
$$=\lim_{h_n \rightarrow 0} \frac{f(a_n+h_n)-f(a_n)}{h_n}$$

The expression is equal to $f'(a_n)$ and $f'(a_n)$ is equal to $f'(x_0)$

By $\epsilon-\delta$, do you mean using $\epsilon-\delta$ definition of limit?

Thanks

 

[PeroK]

Sorry I don't understand the hint.

I followed hint in post#2 and this is what I did:
Let $h_n=b_n -a_n$ and as $n \rightarrow \infty, h_n \rightarrow 0$
$$\lim_{n \rightarrow \infty} \frac{f(b_n)-f(a_n)}{b_n-a_n}$$
$$=\lim_{h_n \rightarrow 0} \frac{f(a_n+h_n)-f(a_n)}{h_n}$$

The expression is equal to $f'(a_n)$ and $f'(a_n)$ is equal to $f'(x_0)$

That needs to be proved.

By $\epsilon-\delta$, do you mean using $\epsilon-\delta$ definition of limit?

Thanks

Yes. What else could that mean?

 

[songoku]

That needs to be proved.

You mean I have to prove $f'(a_n)=f'(x_0)$?

Yes. What else could that mean?

The definition is something that is not taught in class so maybe we are not expected to use that but I will try first and see what I can get from it

Thanks

 

[PeroK]

You mean I have to prove $f'(a_n)=f'(x_0)$?

That statement is clearly not true. You don't even know that for all $n$ $f'(a_n)$ exists.

The definition is something that is not taught in class so maybe we are not expected to use that but I will try first and see what I can get from it

Thanks

Then what definition of a limit are you using?

 

[PeroK]

How about
$$b_n-a_n:=h_n$$
$$h_n\rightarrow 0$$

I'm not convinced this helps. We only know the derivative exists at $x_0$, which is a key point.

 

[PeroK]

@songoku I think you need to expand the expression and pull out a constant term of $f'(x_0)$ and show that what's left converges to zero. There's quite a lot of algebra.

 

[pasmith]

If $f$ is differentiable at $x_0$ then for any $h$, $$f(x_0 + h) = f(x_0) + hf'(x_0) + E(h)$$ where $$\lim_{h \to 0} \frac{E(h)}{h} = 0.$$ If you set $h_n = a_n - x_0$ and $k_n = b_n - x_0$ then the algebra is easier, although I did have to resort to proof by cases depending on whether $|h_n/k_n|$ remains bounded as $n \to \infty$.

 

[anuttarasammyak]

I'm not convinced this helps. We only know the derivative exists at x0, which is a key point.

Let us see it.
$$\lim_{n \rightarrow \infty} \frac{f(a_n+h_n)-f(a_n)}{h_n}$$
If we make it
$$=\lim_{n\rightarrow \infty} \frac{f(x_0+h_n)-f(x_0)}{h_n}$$
It works. Can we make it so ?

 

[PeroK]

Let us see it.
$$\lim_{n \rightarrow \infty} \frac{f(a_n+h_n)-f(a_n)}{h_n}$$
If we make it
$$=\lim_{n\rightarrow \infty} \frac{f(x_0+h_n)-f(x_0)}{h_n}$$
It works. Can we make it so ?

Okay, but you still have to justify that step.

 

[anuttarasammyak]

$$\lim_{n\rightarrow \infty}\frac{f(b_n)-f(a_n)}{b_n-a_n}=\lim_{n\rightarrow \infty}\frac{[f(\beta_n+x_0)-f(x_0)]+[f(x_0)-f(x_0-\alpha_n)]}{\beta_n+\alpha_n}$$
where
$$\beta_n=b_n-x_0$$
$$\alpha_n=x_0-a_n$$, decomposing $h_n$ before. We know
$$\lim_{n\rightarrow \infty}\frac{f(\beta_n+x_0)-f(x_0)}{\beta_n}=\lim_{n\rightarrow \infty}\frac{f(x_0)-f(x_0-\alpha_n)}{\alpha_n}=f'(x_0)$$

 

[PeroK]

$$\lim_{n\rightarrow \infty}\frac{f(b_n)-f(a_n)}{b_n-a_n}=\lim_{n\rightarrow \infty}\frac{[f(\beta_n+x_0)-f(x_0)]+[f(x_0)-f(x_0-\alpha_n)]}{\beta_n+\alpha_n}$$
where
$$\beta_n=b_n-x_0$$
$$\alpha_n=x_0-a_n$$, decomposing $h_n$ before. We know
$$\lim_{n\rightarrow \infty}\frac{f(\beta_n+x_0)-f(x_0)}{\beta_n}=\lim_{n\rightarrow \infty}\frac{f(x_0)-f(x_0-\alpha_n)}{\alpha_n}=f'(x_0)$$

Two things. You have a problem with the denominator and this isn't your homework.

 

[PeroK]

To give a big hint. The first step here is to rewrite the expression as:
$$\frac{f(b_n) - f(a_n)}{b_n - a_n} = \frac{f(b_n) - f(x_0)}{b_n - a_n} + \frac{f(x_0) - f(a_n)}{b_n - a_n}$$$$ = \frac{f(b_n) - f(x_0)}{b_n - x_0}\bigg [\frac{b_n - x_0}{b_n - a_n} \bigg ] + \frac{f(x_0) - f(a_n)}{x_0 - a_n} \bigg [\frac{x_0 - a_n}{b_n - a_n} \bigg ]$$And then to introduce the quantity $f'(x_0)$ ...

 

[anuttarasammyak]

Two things. You have a problem with the denominator and this isn't your homework.

Though this is not my homework let me investigate the problem you find. With $\lim_{n\rightarrow \infty}$, may we equate the fractions as
$$\frac{f(\beta_n+x_0)-f(x_0)}{\beta_n}=\frac{f(x_0)-f(x_0-\alpha_n)}{\alpha_n}=\frac{[f(\beta_n+x_0)-f(x_0)]+[f(x_0)-f(x_0-\alpha_n)]}{\beta_n+\alpha_n}$$?[EDIT] For any n, may we say
$$\min \{\frac{f(\beta_n+x_0)-f(x_0)}{\beta_n},\frac{f(x_0)-f(x_0-\alpha_n)}{\alpha_n}\}$$
$$\leq\frac{[f(\beta_n+x_0)-f(x_0)]+[f(x_0)-f(x_0-\alpha_n)]}{\beta_n+\alpha_n}\leq$$
$$\max \{\frac{f(\beta_n+x_0)-f(x_0)}{\beta_n},\frac{f(x_0)-f(x_0-\alpha_n)}{\alpha_n}\}$$?

 

[songoku]

That statement is clearly not true. You don't even know that for all $n$ $f'(a_n)$ exists.

Ah yes, I see.

Then what definition of a limit are you using?

I am not sure. The teacher said: "If we want to prove limit as x approaches 5 of (x + 1) is 6, then we need to use precise definition of limit but I won't cover that"

So maybe I am not using any definition of limit right now, I am not really sure.

To give a big hint. The first step here is to rewrite the expression as:
$$\frac{f(b_n) - f(a_n)}{b_n - a_n} = \frac{f(b_n) - f(x_0)}{b_n - a_n} + \frac{f(x_0) - f(a_n)}{b_n - a_n}$$$$ = \frac{f(b_n) - f(x_0)}{b_n - x_0}\bigg [\frac{b_n - x_0}{b_n - a_n} \bigg ] + \frac{f(x_0) - f(a_n)}{x_0 - a_n} \bigg [\frac{x_0 - a_n}{b_n - a_n} \bigg ]$$And then to introduce the quantity $f'(x_0)$ ...

I understand this hint

 

[anuttarasammyak]

To give a big hint. The first step here is to rewrite the expression as:
$$\frac{f(b_n) - f(a_n)}{b_n - a_n} = \frac{f(b_n) - f(x_0)}{b_n - a_n} + \frac{f(x_0) - f(a_n)}{b_n - a_n}$$$$ = \frac{f(b_n) - f(x_0)}{b_n - x_0}\bigg [\frac{b_n - x_0}{b_n - a_n} \bigg ] + \frac{f(x_0) - f(a_n)}{x_0 - a_n} \bigg [\frac{x_0 - a_n}{b_n - a_n} \bigg ]$$And then to introduce the quantity $f'(x_0)$ ...

As in my post #2 you leave $b_n-a_n$ in the formula. Now I think it was not a good strategy because of independent convergence of $a_n \rightarrow x_0$ and $b_n \rightarrow x_0$.
The first term of RHS in the limit is
$$f'(x)\lim_{n \rightarrow \infty}(1+\frac{x_0-a_n}{b_n-x_0})^{-1}$$
The second term of RHS in the limit is
$$f'(x)\lim_{n \rightarrow \infty}(1+\frac{b_n-x_0}{x_0-a_n})^{-1}=f'(x)(1-\lim_{n \rightarrow \infty}(1+\frac{x_0-a_n}{b_n-x_0})^{-1})$$
They cancel to bring f'(x) but I wonder according to independent convergence of $a_n \rightarrow x_0$ and $b_n \rightarrow x_0$, though we know that the value for any n is between 0 and 1 anyway, I am not sure there exist the limit value for infinite n. Does the cancellation work even when there exist the no limit value ? I prefer a discussion without comparing the convergences of $a_n$ and $b_n$ .

[EDIT] Now I understand even if there exists no limit, as it is bounded between 0 and 1, we can get the result. Thanks for teachings guys.

 

[PeroK]

The first term of RHS in the limit is
$$f'(x)\lim_{n \rightarrow \infty}(1+\frac{x_0-a_n}{b_n-x_0})^{-1}$$
The first term of RHS in the limit is
$$f'(x)\lim_{n \rightarrow \infty}(1+\frac{b_n-x_0}{x_0-a_n})^{-1}$$

That was not my idea.

 

[songoku]

Thank you very much anuttarasammyak, PeroK, pasmith