Question about differentiability

[pc2-brazil]

I study Calculus by myself, and I tried to solve the following question.
I got the answer, but is my solution consistent?

Thank you in advance.

1. The problem statement
Let f be a function such that |f(x)| ≤ x² for every x. Show that f is differentiable in 0 and that f'(0) = 0.

2. Relevant theorems
I used the Squeeze Theorem, as suggested by the book I'm using. This theorem states the following: "Suppose the functions f, g and h are defined in an open interval I containing a, except possibly at a, and that f(x) ≤ g(x) ≤ h(x) for every x in I such that x ≠ a. If the limits of both f(x) and h(x) as x approaches a exist and equal L, then the limit of g(x) as x approaches a also exists and equals L.".

The Attempt at a Solution

If $$|f(x)| \leq x^2$$, then
$$-x^2 \leq f(x) \leq x^2$$ (1)
Derivative of f at 0:
$$f'(0)=\lim_{h\to 0}\frac{f(0+h) - f(0)}{h}$$
But, from (1), we see that f(0) = 0, so this derivative becomes:
$$f'(0)=\lim_{h\to 0}\frac{f(h)}{h}$$ (2)
From (1), assuming h is approaching zero from positive values:
$$-h^2 \leq f(h) \leq h^2$$
$$-h \leq \frac{f(h)}{h} \leq h$$
From the Squeeze Theorem, we see that, since the limits of $$h$$ and $$-h$$ as h approaches 0 exist and equal 0, then
$$\lim_{h\to 0}\frac{f(h)}{h} = 0$$
Substituting this limit in (2), we see that $$f'(0)=0$$, as desired.

 

[losiu99]

Yep, it's correct.

 

[pc2-brazil]

Yep, it's correct.

Thank you for confirming.