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)
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)
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)
[tex]\lim_{x\to 0}\frac{1-cos(x)}{sin(x)}[/tex]
[tex]\frac{(1-cos(x))(1+cos(x))}{sin(x)(1+cos(x))}[/tex]
[tex]=\frac{1-cos^{2}(x)}{sin(x)(1+cos(x))}[/tex]
The top is equal to [tex]sin^{2}(x)[/tex], so we have:
[tex]\frac{sin^{2}(x)}{sin(x)(1+cos(x))}[/tex]
[tex]=\lim_{x\to 0}\frac{sin(x)}{1+cos(x)}[/tex]
Now, see the limit?. It is easy now.
okay. I see, thanks for the tip .
------------------------------------------------Originally Posted by chao2006
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.
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:
[tex]\lim_{x\to 0}\frac{sin(x)}{x}=1[/tex]
Hello, chao2006!
[tex]\lim_{x\to0} \frac{1-\cos x}{\sin x}[/tex]
[tex]\text{I don't understand why they break up the problem this way: }\:\frac{x}{\sin x}\cdot\frac{1-\cos x}{x}[/tex]
PAULK saw the reason . . .
We're expected to know these two theorems:
. . [tex]\lim_{x\to0}\frac{\sin x}{x} \:=\:1 \qquad\qquad \lim_{x\to0}\frac{1-\cos x}{x} \:=\:0[/tex]
[tex]\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[/tex]
I'm the other of the two guys who "do" homework.
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