# Thread: help with special trig limits

1. ## help with special trig limits

Hi i'm having a problem with this limit. It seems like it woul be simple to do...

using special trig limits find...

lim 1-cosx / x^2
x->0

i tried breaking it up into 1-cosx/x and 1/x but the 1/x is still undefined.

2. $\frac{1-\cos{x}}{x^2} \cdot \frac{1+\cos{x}}{1+\cos{x}} =
$

$\frac{1-\cos^2{x}}{x^2(1 + \cos{x})} =
$

$\frac{\sin^2{x}}{x^2(1 + \cos{x})} =$

$\frac{\sin{x}}{x} \cdot \frac{\sin{x}}{x} \cdot \frac{1}{1+\cos{x}}$

now take the limit.

3. Originally Posted by skeeter
$\frac{1-\cos{x}}{x^2} \cdot \frac{1+\cos{x}}{1+\cos{x}} =
$

$\frac{1-\cos^2{x}}{x^2(1 + \cos{x})} =
$

$\frac{\sin^2{x}}{x^2(1 + \cos{x})} =$

$\frac{\sin{x}}{x} \cdot \frac{\sin{x}}{x} \cdot \frac{1}{1+\cos{x}}$

now take the limit.

Thanks big time! It was that stupid trig identity!! I can never remember those things.

4. Originally Posted by mattboersema
Hi i'm having a problem with this limit. It seems like it woul be simple to do...

using special trig limits find...

lim 1-cosx / x^2
x->0

i tried breaking it up into 1-cosx/x and 1/x but the 1/x is still undefined.

another method is given in the following link which breaks up the
(1-cosx) into half angle
limit

another method could be L'Hospitals rule.

5. according to SANDWICH Theory we can replace

$1-\cos u$

with

$\frac{1}{2} u^2$

when we get ambigious mode of $\frac{0}{0}$

so the menthiond limit can be changed to:

$\lim_ {x\to 0} \frac {\frac {1}{2} x^2}{x^2} = \frac {1}{2}$