Two Trigonometric Limits
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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 nextsmallest) 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.