Show that a limit is 0 if f is differentiable in x=0

[Ozma]

Let [MATH]f:\mathbb{R} \to \mathbb{R}[/MATH] be a function and let [MATH]f[/MATH] differentiable in [MATH]x=0[/MATH]. Show that
[MATH]\lim_{x \to 0} \frac{f(x^2)-f(0)}{x}=0[/MATH]My try: [MATH]f[/MATH] is differentiable in [MATH]x=0[/MATH], so from the Taylor formula with center in [MATH]x=0[/MATH] it is [MATH]f(x)=f(0)+xf'(0)+o(x)[/MATH]; substituting [MATH]x^2[/MATH] in the formula I get [MATH]f(x^2)=f(0)+x^2 f'(0)+o(x^2)[/MATH], so
[MATH]\lim_{x \to 0} \frac{f(x^2)-f(0)}{x}=\lim_{x \to 0} \frac{f(0)+x^2 f'(0)+o(x^2)-f(0)}{x}=\lim_{x \to 0} \frac{x^2 f'(0)+o(x^2)}{x}=\lim_{x \to 0} \left[xf'(0) +\frac{o(x^2)}{x}\right]=0+0=0[/MATH]Where I used the fact that [MATH]f'(0)=l\in\mathbb{R}[/MATH] since by hypothesis [MATH]f[/MATH] is differentiable in [MATH]x=0[/MATH], so [MATH]f'(0)[/MATH] must be finite and real. Is this correct? I'm not sure if the following part is correct: "[MATH]f(x)=f(0)+xf'(0)+o(x)[/MATH]; substituting [MATH]x^2[/MATH] in the formula I get [MATH]f(x^2)=f(0)+x^2 f'(0)+o(x^2)[/MATH]".

 

[pka]

Let [MATH]f:\mathbb{R} \to \mathbb{R}[/MATH] be a function and let [MATH]f[/MATH] differentiable in [MATH]x=0[/MATH]. Show that
[MATH]\lim_{x \to 0} \frac{f(x^2)-f(0)}{x}=0[/MATH]

Because \(f\) is differentiable, it is also continuous at \(x=0\).
Notice that because \(\mathop {\lim }\limits_{x \to 0} f\left( {{x^2}} \right) - f(0) = 0\)
The given limit has the form of \(\dfrac{0}{0}\) so it is l'Hôpital equivlant to \(\mathop {\lim }\limits_{x \to 0} \dfrac{{2xf'\left( {{x^2}} \right)}}{1}=0\)

 

[Ozma]

@pka: Thank you for your approach, it is nice. Can you please tell me if it is correct to substitute [MATH]x^2[/MATH] in the Taylor expansion to get [MATH]f(x^2)=f(0)+x^2 f'(0)+o(x^2)[/MATH]? Have a good day.

 

[Cubist]

Can you please tell me if it is correct to substitute [MATH]x^2[/MATH] in the Taylor expansion to get [MATH]f(x^2)=f(0)+x^2 f'(0)+o(x^2)[/MATH]?

I guess that function o(x) represents:-

• the higher order terms in the Taylor expansion?
• OR, an upper bound error function for a two term Taylor expansion?

Either way, what is your argument that...

[MATH]\lim_{x \to 0} \frac{o(x^2)}{x}=0\,[/MATH] ?

Does o(x) even exist, since f(x) is differentiable (not infinitely differentiable)?

 

[Steven G]

Here is how I would do this problem.

[math]\lim_{x\to 0}\dfrac{f(x^2)-f(0)}{x} = \lim_{x\to 0}x\dfrac{f(x^2)-f(0)}{x^2-0}=0*f'(x^2) = 0.[/math] Note that f'(x²) exists.

 

[Ozma]

@Cubist: Actually it is upper bound error for the first term of Taylor expansion but since I've used [MATH]x^2[/MATH] in place of [MATH]x[/MATH] I've substituted it in the error as well. But I'm still confused about if I can or cannot substitute, in general, [MATH]x^2[/MATH] in place of [MATH]x[/MATH] in a Taylor expansion to get the expression you've quoted.

@Jomo: I have a question, since
[MATH]f'(x_0)=\lim_{x \to x_0} \frac{f(x)-f(x_0)}{x-x_0}[/MATH]Shouldn't it be
[MATH]f'(0)=\lim_{x \to 0} \frac{f(x^2)-f(0)}{x^2-0}[/MATH]Instead of [MATH]f'(x^2)[/MATH]? Or am I wrong?

 

[Steven G]

@Cubist: Actually it is upper bound error for the first term of Taylor expansion but since I've used [MATH]x^2[/MATH] in place of [MATH]x[/MATH] I've substituted it in the error as well. But I'm still confused about if I can or cannot substitute, in general, [MATH]x^2[/MATH] in place of [MATH]x[/MATH] in a Taylor expansion to get the expression you've quoted.

@Jomo: I have a question, since
[MATH]f'(x_0)=\lim_{x \to x_0} \frac{f(x)-f(x_0)}{x-x_0}[/MATH]Shouldn't it be
[MATH]f'(0)=\lim_{x \to 0} \frac{f(x^2)-f(0)}{x^2-0}[/MATH]Instead of [MATH]f'(x^2)[/MATH]? Or am I wrong?

Yes, I am guilty of sloppy thinking. Good job finding my mistake

 

[Cubist]

But I'm still confused about if I can or cannot substitute, in general, [MATH]x^2[/MATH] in place of [MATH]x[/MATH] in a Taylor expansion to get the expression you've quoted.

In general you can consider a Taylor series like any other function, and you can pass in any parameter that you want to.

Let's consider a general function g(x), and it's Taylor series exists and is t(x).
Let h(x)=x^2. Another Taylor series could be found for g(h(x)). Lets call this series t₂(x).
Then, t₂(2)≈g(4)≈t(4). Or more generally t₂(x)≈g(x^2)≈t(x^2)

I use "approx equal to" symbol assuming that all parameters are within reasonable range of the point for which the two Taylor series were calculated. I hope this helps, and does not confuse you even more!

--

...but as I said before, the method in the OP does not provide a definition of o(x), nor consider if a definition is possible, therefore I think it is probably a flawed method for that reason.

 

[Ozma]

@Cubist: Thank you for your clarification, I appreciate it. Ok, sorry for not defining the meaning of [MATH]o(x)[/MATH]: I mean that [MATH]f[/MATH] is an [MATH]o(x)[/MATH] as [MATH]x \to x_0[/MATH] if
[MATH]\lim_{x \to x_0} \frac{f(x)}{x}=0[/MATH]So, with this definition, [MATH]\frac{o(x^2)}{x} \to 0[/MATH] as [MATH]x \to 0[/MATH]; but I'm not sure if you mean that I cannot write [MATH]o(x^2)[/MATH] since I don't have enough regularity of [MATH]f[/MATH] in the hypothesis of the problem and so my initial approach is wrong because of this.

 

[Cubist]

@Cubist: Thank you for your clarification, I appreciate it.

You're welcome!

I mean that [MATH]f[/MATH] is an [MATH]o(x)[/MATH] as [MATH]x \to x_0[/MATH] if
[MATH]\lim_{x \to x_0} \frac{f(x)}{x}=0[/MATH]

I'm only familiar with the following method of finding an upper bound error function - https://brilliant.org/wiki/taylor-series-error-bounds/ . This method would require the second derivative (in your proof). But we have not been told that a second derivative exists for the given function, therefore this method could not be used for this question.

PERHAPS what you have stated in the above quote is a standard/known property of the upper bound error function - but I can't verify this myself (I have not seen it before).

 

[Cubist]

PERHAPS ...

I just re-read my post and it seemed a bit sarcastic with that capitalization, sorry. I actually just wanted to emphasise that you might be correct, but I can't verify it.

 

[Ozma]

@Cubist: It's okay, I didn't think it was sarcastic; but I like the fact that you thought about that point of view.

I actually just wanted to emphasise that you might be correct, but I can't verify it.

Thanks, I will keep it in mind. If you want to answer in future about this, it is welcome.