 ## Two Trigonometric Limits Español

We will show the two results

 1. $\lim_{h \to 0} \frac{\sin h}{h} = 1$      2. $\lim_{h \to 0} \frac{\cos h - 1}{h} = 0$

Proof of (1)

First take a look at the following diagram, showing three areas arranged in order of magnitude. The shaded area on the left (the smallest of the three) is a triangle with height of length $\sin h$ and base of length $\cos h.$ Therefore, its area is $\frac{(\cos h)(\sin h)}{2}.$

The pink shaded area (the next-smallest) is a circular segment compromising the fraction $\frac{h}{2π}$ of the entire disc. Since the area of a disc of radius $1$ is $π,$ the area in question is

$\frac{h}{2π}\ .\ π = \frac{h}{2}$

The shaded area on the right (the largest of the three) is a triangle with height of length $\tan h$ and base of length $1.$ Therefore, its area is $\frac{(1)(\tan h)}{2} = \frac{\tan h}{2}.$

Substituting these three areas therefore gives the inequality

$\frac{\cos h\ \ \sin h}{2}\ ≤\ \frac{h}{2}\ ≤\ \frac{\tan h}{2}.$

Writing $\tan h$ as the ratio $\frac{\sin h}{\cos h}$ now gives

$\frac{\cos h\ \ \sin h}{2}\ ≤\ \frac{h}{2}\ ≤\ \frac{\sin h}{2\cos h}.$

Multiplying through by $\frac{2}{\sin h}$ now gives

$\cos h\ ≤\ \frac{h}{\sin h}\ ≤\ \frac{1}{\cos h}.$

Now take reciprocals and reverse inequalities to get

$\frac{1}{\cos h}\ ≥\ \frac{\sin h}{h}\ ≥\ \cos h.$

Finally, let $h$ approach zero. As is does, the quantities on either end approach $1.$ Therefore, since the ratio $\frac{\sin h}{h}$ is sandwiched between two quantities approaching $1,$ it also approaches $1.$

We are now done with the first limit we promised to compute.

Proof of (2)

For the second limit, we use a trigonometric identity and a little algebra:

 $\lim_{h \to 0} \frac{1 - \cos h}{h}$ $=$ $\lim_{h \to 0} \frac{1 - \cos h}{h}\ .\ \frac{1 + \cos h}{1 + \cos h}$ $=$ $\lim_{h \to 0} \frac{1 - \cos^2 h}{h(1 + \cos h)}$ $=$ $\lim_{h \to 0} \frac{\sin^2 h}{h(1 + \cos h)}$ (using the identity $\sin^2 h + \cos^2 h = 1)$ $=$ $\lim_{h \to 0} \frac{\sin h}{h}\ .\ \frac{\sin h}{1 + \cos h}$

The first term in this product is the limit we computed above, and has the value of $1.$ The second term approaches $\frac{0}{(1+1)} = 0.$ Therefore, the product approaches $(1)(0) = 0,$ as required.  