# Thread: limits: the limit, as x -> 0, of (1 - cosx) / sinx

1. ## limits: the limit, as x -> 0, of (1 - cosx) / sinx

I have trouble solving trig limits...

lim
x--> 0
1-cosx
sinx

I don't understand why they break up the problem this way... to this:

(x) (1-cosx)
(sinx) (x)

2. ## Re: limits

Here's one way to do it. Remember this method. It will come in handy with limits. That way you can transform it into something you can work with.

Multiply top and bottom by 1+cos(x)

$\lim_{x\to 0}\frac{1-cos(x)}{sin(x)}$

$\frac{(1-cos(x))(1+cos(x))}{sin(x)(1+cos(x))}$

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

The top is equal to $sin^{2}(x)$, so we have:

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

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

Now, see the limit?. It is easy now.

3. ## Re: limits

okay. I see, thanks for the tip .

4. ## Re: limits: the limit, as x -> 0, of (1 - cosx) / sinx

Originally Posted by chao2006
I have trouble solving trig limits...

lim
x--> 0
1-cosx
sinx

I don't understand why they break up the problem this way... to this:

(x) (1-cosx)
(sinx) (x)
------------------------------------------------
Sorry about the mangling of your question -- it's a total mess and the SYSTEM did that to us.
Anyway, I think 'they' wanted to test your knowledge of these 'standard' limits:

sin x
----- --> 1
x
1 - cos x
------------ --> 0 << typo fixed.
x
Of course the example can be done without putting in the x's, as the last post showed.

5. ## Re: limits: the limit, as x -> 0, of (1 - cosx) / sinx

That's why we have LaTex. Or you can wrap it in 'code' tags to keep it in line.

Code:
                  sinx
lim        ------  = 1
x->0            x

This looks better though:

$\lim_{x\to 0}\frac{sin(x)}{x}=1$

6. ## Re: limits: the limit, as x -> 0, of (1 - cosx) / sinx

Hello, chao2006!

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

$\text{I don&#39;t understand why they break up the problem this way: }\:\frac{x}{\sin x}\cdot\frac{1-\cos x}{x}$

PAULK saw the reason . . .

We're expected to know these two theorems:

. . $\lim_{x\to0}\frac{\sin x}{x} \:=\:1 \qquad\qquad \lim_{x\to0}\frac{1-\cos x}{x} \:=\:0$

$\text{Then: }\;\lim_{x\to0}\frac{1-\cos x}{\sin x} \;=\;\lim_{x\to0}\left[\frac{x}{\sin x}\cdot\frac{1-\cos x}{x}\right] \;=\;1\cdot 0 \;=\;0$

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