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limits: the limit, as x -> 0, of (1 - cosx) / sinx
- Thread starter chao2006
- Start date
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)
\(\displaystyle \lim_{x\to 0}\frac{1-cos(x)}{sin(x)}\)
\(\displaystyle \frac{(1-cos(x))(1+cos(x))}{sin(x)(1+cos(x))}\)
\(\displaystyle =\frac{1-cos^{2}(x)}{sin(x)(1+cos(x))}\)
The top is equal to \(\displaystyle sin^{2}(x)\), so we have:
\(\displaystyle \frac{sin^{2}(x)}{sin(x)(1+cos(x))}\)
\(\displaystyle =\lim_{x\to 0}\frac{sin(x)}{1+cos(x)}\)
Now, see the limit?. It is easy now.
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)
\(\displaystyle \lim_{x\to 0}\frac{1-cos(x)}{sin(x)}\)
\(\displaystyle \frac{(1-cos(x))(1+cos(x))}{sin(x)(1+cos(x))}\)
\(\displaystyle =\frac{1-cos^{2}(x)}{sin(x)(1+cos(x))}\)
The top is equal to \(\displaystyle sin^{2}(x)\), so we have:
\(\displaystyle \frac{sin^{2}(x)}{sin(x)(1+cos(x))}\)
\(\displaystyle =\lim_{x\to 0}\frac{sin(x)}{1+cos(x)}\)
Now, see the limit?. It is easy now.
------------------------------------------------chao2006 said: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.
Hello, chao2006!
PAULK saw the reason . . .
We're expected to know these two theorems:
. . \(\displaystyle \lim_{x\to0}\frac{\sin x}{x} \:=\:1 \qquad\qquad \lim_{x\to0}\frac{1-\cos x}{x} \:=\:0\)
\(\displaystyle \text{Then: }\;\lim_{x\to0}\frac{1-\cos x}{\sin x} \;=\;\lim_{x\to0}\left[\frac{x}{\sin x}\cdot\frac{1-\cos x}{x}\right] \;=\;1\cdot 0 \;=\;0\)
\(\displaystyle \lim_{x\to0} \frac{1-\cos x}{\sin x}\)
\(\displaystyle \text{I don't understand why they break up the problem this way: }\:\frac{x}{\sin x}\cdot\frac{1-\cos x}{x}\)
PAULK saw the reason . . .
We're expected to know these two theorems:
. . \(\displaystyle \lim_{x\to0}\frac{\sin x}{x} \:=\:1 \qquad\qquad \lim_{x\to0}\frac{1-\cos x}{x} \:=\:0\)
\(\displaystyle \text{Then: }\;\lim_{x\to0}\frac{1-\cos x}{\sin x} \;=\;\lim_{x\to0}\left[\frac{x}{\sin x}\cdot\frac{1-\cos x}{x}\right] \;=\;1\cdot 0 \;=\;0\)