Let $f$ be a real valued differentiable function on the real line such that $\lim_{x\to 0}\frac{f(x)}{x^2}$ exists, and is finite . Prove $f'(0)=0.$

Question

Let $f$ be a real valued differentiable function on the real line $\Bbb R$ such that $\lim_{x\to 0}\frac{f(x)}{x^2}$ exists, and is finite . Prove that $f'(0)=0$

My solution goes like this:

Given, $$f'(0)=\lim_{x\to 0}\frac{f(x)}{x^2}.$$ Now, we have $f$ differentiable over the real line, this implies that $f$ is differentiable everywhere. Now, as $f$ is differentiable everywhere, this means that that $f$ is continuous everywhere as well. If $f(0)=r$ such that $r\neq 0$ then if $\lim_{x\to 0}\frac{f(x)}{x^2}=\frac{f(0)}{0}=L$ then $L=\infty$ but this is not possible as it is mentioned in the question that, $L$ is finite. So, we can conclude $f(0)=0.$ We know that $$f'(0)=\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to 0}\frac{f(x)}{x}=\lim_{x\to 0}\frac{xf(x)}{x^2}=0\times L=0.$$ This completes the proof.

However, I feel unsure that whether my "process" or rather my way of conccluding $f(0)=0$ is justified or not? I get somewhat of an uncanny feeling from this part of my proof. Iff that is valid, then is my proof a completely justified approach?

Answer 1

I don't know if this is the correct solution, but give a counterexample if it is wrong.

Let $f$ be a real valued differentiable function on the real line $\mathbb R$ such that $\lim_{x \to 0} \frac{f(x)}{x^2}$ exists, and is finite. Prove that f′(0)=0.

For the given limit to exsist (I think) that the function $f(x)$ has to be a polynomial of degree higher than $n=2$ so the terms can cancel and the limit will not blow up because of the denominator.

So for any polynomial $p(x) = a x^n + b x^{n-1} + \cdots$ , $n>2$ the limit will exsist.

Now to show that $f(0)' = 0$, we can see that the last term in the polynomial (the one without x will be $0$ after derivation and all the other terms with $x$ in them will be $0$ when we input $0$ as we have to. So that is it.

$f(0)' = 0$ and the limit $\lim_{x \to 0} \frac{f(x)}{x^2}$ exists.

Answer 2

There is a simpler, more rigorous way to conclude $f(0) = 0$: suppose $\lim_{x\to0} f(x)/x^2 = L$, then because $f$ is continuous and $L$ is finite, $$ f(0) = \lim_{x\to0} f(x) = \lim_{x\to0}\frac{x^2f(x)}{x^2} = \left(\lim_{x\to0} x^2\right)\left(\lim_{x\to0}\frac{f(x)}{x^2}\right) = 0\cdot L = 0. $$ The rest of your solution looks correct from there. Fun fact, if $f''(0)$ exists, then the limit in question must be $L = \frac12f''(0)$ due to L'Hospital's rule and the fact that $f(0) = f'(0) = 0$.