$$ f(x) = \begin{cases} \frac{\sin x}{x}, & \quad x \neq 0 \\ 1, & \quad x = 0 \end{cases} $$

Prove that $f$ is differentiable at $0$ and find $f'(0)$.


I'm working through Spivak's Calculus and this question is being asked to me after I covered the Fundamental of Calculus and defining the $\sin$ and $\cos$ functions formally. I don't think it has much to do with that, but just as a caveat.

To prove the result I resorted to applying the definition that a function is differentiable at a point $c$. That is

$f$ is differentiable at the point $0$ if $\lim_{h \to 0} \frac{f(0 + h) - f(0)}{h} = \text{some value}$.

Using this idea and some algebra I arrive at:

$$\lim_{h \to 0} \frac{\frac{\sin(h)}{h} - 1}{h} = \lim_{h \to 0} \frac{\sin(h) - h}{h^{2}}$$

I was expecting some simple cancellation of $h$ through out the expression, but alas that did not occur. I did think of using the idea that:

$$-\frac{1}{h} \leq \frac{\sin(h)}{h} \leq \frac{1}{h}$$,

but I don't see much coming from it. What step am I missing?

  • $\begingroup$ How are sine and cosine defined? Are you allowed to use the Taylor series? $\endgroup$ – Brian Tung Aug 29 '20 at 22:32
  • $\begingroup$ You have a typo in your last inequalities. Note that Spivak specifically tells you to use L'Hôpital's rule. $\endgroup$ – Ted Shifrin Aug 29 '20 at 22:36
  • $\begingroup$ @TedShifrin, didn't even know this was specifically a Spivak question from the text. I took it from a set of handouts that I'm working along with in companion to the text. Do you know what question it is specifically in the text? $\endgroup$ – dc3rd Aug 29 '20 at 22:42
  • 1
    $\begingroup$ In the third/fourth edition, #3 in chapter 15. $\endgroup$ – Ted Shifrin Aug 29 '20 at 23:39

We can use l'Hospital to obtain

$$\lim_{h \to 0} \frac{\sin(h) - h}{h^{2}}=\lim_{h \to 0} \frac{\cos(h) - 1}{2h}=0$$

indeed by definition of derivative

$$\lim_{h \to 0} \frac12\frac{\cos(h) - 1}{h}=-\frac12\sin (0)=0$$

  • $\begingroup$ Why did I not see the glaring L'Hopital, especially considering that the first part of the question asked me to shoe $f(x)$ is continuous and I used it then..... Thank you for the help. $\endgroup$ – dc3rd Aug 29 '20 at 22:35
  • $\begingroup$ @dc3rd Yes in this case we need that in order to avoid more complicated ways! You are welcome. Bye $\endgroup$ – user Aug 29 '20 at 22:36

Using Taylor-Young expansion of $ \sin(h) $ around $ h=0 $, we get

$$\sin(h)=h-\frac{h^3}{6}(1+\epsilon(h))$$ with $$\lim_{h\to 0}\epsilon(h)=0$$


$$\lim_{h\to 0}\frac{\sin(h)-h}{h^2}=\lim_{h\to 0}-\frac{h}{6}(1+\epsilon(h))=0$$

So, $ f $ is differentiable at $ x=0$ and $f'(0)=0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.