1. ## limit of trigo

how do you solve lim ((cos x - 1) / tan x) as x -> 0?

i tried to express it as lim ((cos ^2 x - cos x)/ sinx) = (1-1)/0 = undefined but according to my notes, the answer is 0

2. $\displaystyle \lim_{x\to0}\frac{\cos x - 1}{\tan x}$

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

$\displaystyle \frac{0}{\infty} - \frac{1}{\infty}$

$\displaystyle 0$

bro.

3. Do you know how to use L'Hospital rule?

You have the form $\displaystyle \frac{0}{0}$

4. Originally Posted by Skerven
$\displaystyle \lim_{x\to0}\frac{\cos x - 1}{\tan x}$

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

$\displaystyle \frac{0}{\infty} - \frac{1}{\infty}$

$\displaystyle 0$

bro.
This is not correct.

$\displaystyle \lim_{x\to0}\frac{\cos x}{\tan x}\to\frac{1}{0}$
and
$\displaystyle \lim_{x\to0}\frac{1}{\tan x}\to\frac{1}{0}$

So you have to use some other method.

5. im not sure about that. how do you use L'Hospital rule?

6. You do the derivative or the top over the derivative of the bottom.

7. Hello, alexandrabel90!

How do you solve: .$\displaystyle \lim_{x\to0}\frac{\cos x - 1}{\tan x}$

We have: .$\displaystyle \frac{-(1-\cos x)}{\frac{\sin x}{\cos x}} \;=\;-\frac{\cos x(1-\cos x)}{\sin x}$

Multiply by $\displaystyle \frac{1+\cos x}{1+\cos x}$

. . $\displaystyle -\frac{\cos x(1 - \cos x)}{\sin x}\cdot\frac{1+\cos x}{1 + \cos x} \;\;=\;\;-\frac{\cos x(1-\cos^2\!x)}{\sin x(1+\cos x)} \;\;=\;\;-\frac{\cos x\sin^2\!x}{\sin x(1 + \cos x)}$ . $\displaystyle = \;\;-\frac{\cos x \sin x}{1 + \cos x}$

Then: .$\displaystyle \lim_{x\to0}\left[-\frac{\cos x\sin x}{1 + \cos x}\right] \;\;=\;\; -\frac{1\cdot0}{1 + 1} \;\;=\;\;-\frac{0}{2} \;\;=\;\;0$