Trigonometric functions.

Proposition 4.5.6       


$\displaystyle \underset{x \rightarrow 0}{\lim} \sin x = 0.$    


$\displaystyle \underset{x \rightarrow 0}{\lim} \cos x = 1.$    


$\displaystyle \underset{x \rightarrow 0}{\lim} \tan x = 0.$    

The following proposition has a nice geometric proof, using the sandwich theorem (later on, 6.1). A more efficient way to prove it is to use either L'Hopital's rule ( i.e. Theorem L'Hopital) or MacLaurin polynomials (v.i. def Taylor polynomial); we will learn them later.

Proposition 4.5.7       


$\displaystyle \underset{x \rightarrow 0}{\lim} \frac {\sin x}{x} = 1.$    


$\displaystyle \underset{x \rightarrow 0}{\lim} \frac {\tan x}{x} = 1.$    


$\displaystyle \underset{x \rightarrow 0}{\lim} \frac {\cos x-1}{x^2}=-\frac 12.$    

Example 4.5.8   Let $ f(x)= \frac {\sin 3x}{\tan 5x}$ . Compute the limit, if it exists, of $ f$ for $ x$ arbitrary close to 0.

$\displaystyle \frac {\sin 3x}{\tan 5x} = \frac {\sin 3x}{3x} \cdot \frac {5x}{\tan 5x} \cdot \frac {3x}{5x}.$    

As $ \underset{x \rightarrow 0}{\lim} 3x = \underset{x
\rightarrow 0}{\lim} 5x = 0$ , we have:

$\displaystyle \underset{x \rightarrow 0}{\lim} \frac {\sin 3x}{\tan 5x} = \frac 35.$    

Note that there are other ways to solve this question; one of them will appear in chapter 6, section section l'hopital, another one in chapter 10, section v6.

Noah Dana-Picard 2007-12-28