1. ## Trig Limits

Hi
I'm having serious trouble with trig limits and I'm wondering if anyone could explain it to me in simple terms as I don't really understand the textbook I have. Any help would be greatly appreciated.

For example in my textbook there is an example however I don't understand it. Could any help and just talk us through it?

lim x approaches 0 (1 - cos x)/x = lim x app 0 ((1 - cos x) / x (1 + cos x)/ 1 + cos x)
I don't understand where the 1 + cos x divided by x came from
then it goes on
= 1 - cos^2 x / x(1 + cos x)
= sin^2 x/ x (1 + cos x)
= (sin x/x * sinx x /(1 + cos x))
= sinx x * lim x app 0 sin x/(1 + cos x)
= 1 * sin 0/1 + cos 0
= 0

I can't follow it through so if anyone can explain the steps and tell me how you get that it would ease my troubles. Thanks.

2. $\lim_{x\to 0} \frac{1-\cos x}{x}$

$\lim_{x\to 0} \frac{1-\cos x}{x}\cdot \frac{1+\cos x}{1+\cos x}$

$\lim_{x\to 0} \frac{1-\cos^2 x}{x(1 + \cos x)}$

$\lim_{x\to 0} \frac{\sin^2 x}{x(1+\cos x)}$

As $\lim_{x\to 0} \frac{\sin x}{x} = 1$,

$\lim_{x\to 0} \frac{\sin^{\not 2} x}{\not{x}(1+\cos x)}$

$\lim_{x\to 0} \frac{\sin x}{1+\cos x}$

Now put x=0,

$\frac{\sin 0}{1+\cos 0}$

$\frac{0}{2} = 0$

That $\frac{1+\cos x}{1+\cos x}$ didn't come from anywhere, we multiplied $\frac{1-\cos x}{x}$ by it to get a solution. To see such things you have to be familiar with trigonometric limits and identities.

3. Hello, Mickey!

A slightly different explanation . . .

We're expected to know that: . $\lim_{x\to0}\frac{\sin x}{x} \:=\:1$

$\lim_{x\to0} \frac{1-\cos x}{x}$

Multiply by $\frac{1+\cos x}{1 + \cos x}\!:\;\;\;\frac{1-\cos x}{x}\cdot{\color{blue}\frac{1+\cos x}{1 + \cos x}} \;\;=\;\;\frac{1-\cos^2x}{x(1+\cos x)} \;\;=\;\;\frac{\sin^2x}{x(1+\cos x)}$

And we have: . $\lim_{x\to0}\left[\frac{\sin x}{x}\cdot\frac{\sin x}{1 + \cos x}\right] \;\;=\;\;1\cdot\frac{0}{1+1} \;\;=\;\;0$