0
$\begingroup$

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

  • 0
    are you allowed to use l'Hopital? – 2017-01-27
  • 0
    I forgot to mention that we neither covered integral calculus nor l´hopital´s rule – 2017-01-27
  • 0
    I added an alternate solution. Should I expand more detail? – 2017-01-27

2 Answers 2

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 $f(x)=x^2.$

  • 0
    Thank you for your advice. So since $g(x)$ is continuous and differentiable for all $x$ and $f$ is continuous and differentiable at all $g(x)$ it follows that $f(g(x))$ is continous and differentiable at all $x$ including $0$. Is that right? – 2017-01-27
  • 0
    Yes. But we still have to show that $\sin x/x$ is differentiable. – 2017-01-27
  • 0
    Much appreciated! The differentiability of $sin x/x$ for all $x$ is clear to me! – 2017-01-27
  • 0
    Really? The continuity of $\sin x/x$ follows from the special limit, but its differentiability at 0 is not at all clear to me without using l'hopital or Taylor series. – 2017-01-27
  • 0
    Yes, with Taylor series it is clear. I just wasn't sure it was allowed. If so, then we're good! – 2017-01-27
  • 0
    btw, in regards to your $g(x)$ function. Shouldn´t it be defined $1$ for $x=0$ ? – 2017-01-27
  • 0
    Oops, yes of course. I will fix. Thank you for the correction. – 2017-01-27
0

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 $(1- 1)(1+ 1)= 0$.

  • 0
    what if we want the limit of $(\sin^2h-h^2)/h^3$ instead? – 2017-01-27