Two important Limits

What we need

To figure out the derivatives of trig functions we need:

Two extremely important limits (derived below): $$ \lim_{x \to 0} \frac{\sin(x)}{x} = 1 \qquad \mbox{and} \qquad \lim_{x \to 0}\frac{1-\cos(x)}{x}=0;$$ The addition-of-angle formulas for sine and cosine: \begin{eqnarray*}\sin(A+B) &=& \sin(A)\cos(B) + \cos(A)\sin(B), \cr \cos(A+B) & = & \cos(A)\cos(B) - \sin(A)\sin(B), \end{eqnarray*} which we covered earlier (review). The quotient rule, and The Pythagorean identities: \begin{eqnarray*} \sin^2(x)+\cos^2(x)=1, \cr \tan^2(x) + 1 = \sec^2(x), \cr 1 + \cot^2(x) = \csc^2(x). \end{eqnarray*}

The Two important Limits

In the following video, we use the squeeze theorem to derive the first of the two limits:

$$ \lim_{x \to 0} \frac{\sin(x)}{x} = 1$$

Once we know the first limit, the second limit

$$ \lim_{x \to 0}\frac{1-\cos(x)}{x}=0$$

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