differentiability of a function at a point containing sine

Question

I want to determine whether the following function is differentiable at $x=0$:

$f: \mathbb{R} \rightarrow \mathbb{R}$, $x \mapsto \frac{{\sin}^2(x)}{x^2}$ for $x\neq 0$ and $1$ for $x=0$.

Clearly, this function is continuous $\forall x \in \mathbb{R}$ and the derivative exists $\forall x \in \mathbb{R}\backslash\{0\}$. I assume that this function is differentiable at $x=0$ with the derivative $ \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} =0$.

By using the definition of the derivative I come to the point where I´m stuck with the following expression: $\lim_{h\to 0}\frac{\frac{{\sin}^2(h)}{h^2}-1}{h} = \lim_{h\to 0}\frac{{\sin}^2(h)-h^2}{h^3}$. I know that $lim_{h\to 0}\frac{\sin(h)}{h} = 1$ and my guess is that I have to work with $g(x):= \lim_{h\to 0}\frac{\sin(h)}{h}$ and $f(x):=x^2$, so $f(g(x)) = f(1) = 1$. Any help would be appreciated.

Thank you, Hofmusicus

Answer 1

Using l'Hôpital, the limit can be computed

$$ \lim_{h\to 0} \frac{\sin^2h-h^2}{h^3} = \lim_{h\to 0} \frac{2\sin h\cos h-2h}{3h^2} = \lim_{h\to 0} \frac{2\cos 2h-2}{6h} = \lim_{h\to 0} \frac{-4\sin 2h}{6}=0 $$.

Another alternative perhaps more in line with your thinking, is to consider the function

$$ g(x) = \operatorname{sinc} x = \left\{ \begin{array}{l} &\frac{\sin x}{x} & x\neq 0\\ &1 &x=0 \end{array} \right. $$ Check that $g(x)$ is continuous and differentiable at 0 using the $\frac{\sin x}{x}\to 1$ limit that you know, and then by general properties of composite functions $f(g(x))$ is also continuous and differentiable, with $f(x)=x^2.$

Answer 2

Use $a^2- b^2= (a- b)(a+ b)$ to factor $sin^2(h)- h^2$ as $(sin(h)- h)(sin(h)+ h)$ so that $\frac{sin^2(h)- h^2}{h^2}= \left(\frac{sin(h)- h}{h}\right)\left(\frac{sin(h)+ h)}{h}\right)$$= \left(\frac{sin(h)}{h}- 1\right)\left(\frac{sin(h)}{h}+ 1\right)$. Since the limit of $\frac{sin(h)}{h}$ is 1, that becomes $(1- 1)(1+ 1)= 0$.