Which of the following about $f(x)$ is correct?

Question

Let $f:\mathbb R \to \mathbb R$ be defined by $$f(x) = \left\{ \begin{array}{1l} \dfrac{\sin x}{x} & x \neq 0 \\ 1 & x = 0 \\ \end{array} \right.$$ Then:

1. $f$ is not continuous.

2. $f$ is continuous but not differentiable.

3. $f$ is differentiable

Solution: Clearly $f(x)$ is continuous at $x=0$ and $\lim_{x\to 0^{+}}\frac{\sin x}{x}=1=\lim_{x\to 0^{-}}\frac{\sin x}{x}$.

Now, about differentiability at $x=0$, $\lim_{x\to 0^{+}}\frac{\frac{\sin h}{h}-1}{h}=\frac{\sin h}{h^2}-\frac{1}{h}$ and $\lim_{x\to 0^{-}}\frac{1-\frac{\sin (-h)}{-h}}{h}=-\frac{\sin h}{h^2}+\frac{1}{h}$ . Now both these limits should go to $0$ as $f'(0)=0$. But they are not going to zero. Hence, the function is continuous but not differentiable.

Answer 1

:Option (3) is correct

$\lim_{h\to 0^{+}}\frac{\frac{\sin h}{h}-1}{h}$ $(\frac{0}{0}) $ form. Now use L'Hospital's rule we get $\lim_{h \to 0^{+}}\frac{\frac{h\cos h-\sin h}{h^2}}{1} =\lim_{h \to 0^{+}}\frac{h\cos h-\sin h}{h^2}(\frac{0}{0})form$

again using L'Hospital's rule, $\lim_{h \to 0^+}(-\frac{sinh}{2})=0$

Similarly using L'Hospital's rule $ \lim_{h\to 0^{-}}\frac{\frac{\sin h}{h}-1}{h}(\frac{0}{0} form)=0$ Thus $$\lim_{h\to 0}\frac{\frac{\sin h}{h}-1}{h}=0$$ Thus when $x\ne 0,f'(x)=\frac{x\cos {x}-\sin {x}}{x^2}$ And when $x=0 ,f'(x)=0$

Answer 2

The function is differentiable.

$f'(0) $ is not $0$ as you mentioned.

Perhaps you think that it is constant at $0$ and so derivative of a constant function is $0$, that is not the case. Apply the definitions carefully and see this for yourself. Remember, the function defined in the neighbourhood matters.

You will later learn (or might have already learnt) that in order to define derivative you need the function defined in an open neighbourhood. (I am not sure if this makes sense to you.)

Answer 3

Actually $f$ is infinitely differentiable

Your limit is right $ \dfrac{\sin(h)}{h^2}=\frac{h-h^3/6}{h^2}+o(h) $ developing $\sin$ asyptotically at second order so $\dfrac{\sin(h)}{h^2}-\dfrac{1}{h}=-h/6+o(h)$ Actually you're dealing with $sinc$ (say cardinal sinus function). $f$ would by infinitely differentiable. In order to show it you can develop it into a serie.

$$ \forall x\in \mathbb{R}\ , \ f(x)=\sum_{n=0}^\infty \dfrac{(-1)^nx^{2n}}{(2n+1)!}$$

so $f$ is infinitely differentiable on its circle of convergence (here $\mathbb{R}$ because the radius of convergence equal to $\infty$)