Limits necessary to determine the derivatives of $\sin x$ and $\cos x$

On this page, we set the stage to derive the values of $(\sin x)'$ and $(\cos x)'$.  This is particularly interesting to those of you who are curious about how we determine such derivatives for which we do not (yet) have a derivative rule.

You will need to know the values of the following limits later in the semester, not only for this application. 

Unless you are told otherwise, you will not be expected to be able to reproduce the following proofs, but it is certainly true that working through the rest of this page, carefully, will increase your mathematical understanding in many ways.

The two important limits (derived below): $$ \lim_{x \to 0} \frac{\sin(x)}{x} = 1 \qquad \mbox{and} \qquad \lim_{x \to 0}\frac{1-\cos(x)}{x}=0.$$ Notice that both of these limits have the $\frac{0}{0}$ indeterminate form, so we need to do more work, but what work?

is fairly easy: \begin{eqnarray*} \displaystyle{\lim_{x \to 0} \frac{\cos(x)-1}{x}} & = & \displaystyle{\lim_{x \to 0} \left ( \frac{\cos(x)-1}{x}\cdot \frac{\cos(x)+1}{\cos(x)+1} \right ) } \cr & = & \displaystyle{\lim_{x \to 0} \left ( \frac{\cos^2(x)-1}{x(1+\cos(x))}\right )} \cr &=&\displaystyle{\lim_{x \to 0} \left (\frac{-\sin^2(x)}{x(1+\cos(x))}\right )} \cr &=& \displaystyle{-\lim_{x \to 0} \left ( \frac{\sin(x)}{x}\cdot\frac{\sin(x)}{(1+\cos(x))}\right )}\cr &=& \displaystyle{-\left ( \lim_{x \to 0} \frac{\sin(x)}{x} \right )} \cdot \displaystyle{\left ( \lim_{x \to 0} \frac{\sin(x)}{1+\cos(x)}\right )}. \end{eqnarray*}

We just showed that the first of the limits on the last line is 1, and the second limit is $\displaystyle{\frac{\sin(0)}{1+\cos(0)}=\frac{0}{2}=0}$, so the product is 0.