# Prove $\frac{\sin x}{x}$ is continuous at $0$ - proof assistance

Define

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

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

Attempt

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?

• How are sine and cosine defined? Are you allowed to use the Taylor series? – Brian Tung Aug 29 '20 at 22:32
• You have a typo in your last inequalities. Note that Spivak specifically tells you to use L'Hôpital's rule. – Ted Shifrin Aug 29 '20 at 22:36
• @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? – dc3rd Aug 29 '20 at 22:42
• In the third/fourth edition, #3 in chapter 15. – 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

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

• 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. – dc3rd Aug 29 '20 at 22:35
• @dc3rd Yes in this case we need that in order to avoid more complicated ways! You are welcome. Bye – 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 $$\lim_{h\to 0}\epsilon(h)=0$$

thus

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

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