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Math Help - limit of trigo

  1. #1
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    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
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  2. #2
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    <br />
\lim_{x\to0}\frac{\cos x - 1}{\tan x}<br />

    <br />
\lim_{x\to0}\frac{\cos x}{\tan x} - \lim_{x\to0}\frac{1}{\tan x}<br />

    <br />
\frac{0}{\infty} - \frac{1}{\infty}<br />

    <br />
0<br />

    bro.
    Last edited by Skerven; November 9th 2009 at 04:01 PM.
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  3. #3
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    Do you know how to use L'Hospital rule?

    You have the form \frac{0}{0}
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  4. #4
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    Quote Originally Posted by Skerven View Post
    <br />
\lim_{x\to0}\frac{\cos x - 1}{\tan x}<br />

    <br />
\lim_{x\to0}\frac{\cos x}{\tan x} - \lim_{x\to0}\frac{1}{\tan x}<br />

    <br />
\frac{0}{\infty} - \frac{1}{\infty}<br />

    <br />
0<br />

    bro.
    This is not correct.

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

    So you have to use some other method.
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  5. #5
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    im not sure about that. how do you use L'Hospital rule?
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  6. #6
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    You do the derivative or the top over the derivative of the bottom.
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  7. #7
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    Hello, alexandrabel90!

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

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


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

    . . -\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)} . = \;\;-\frac{\cos x \sin x}{1 + \cos x}


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

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