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Math Help - find limit as x approach 0 : [cos(2x) -1]/sin7x

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    find limit as x approach 0 : [cos(2x) -1]/sin7x

    find limit as x approach 0 : [cos(2x) -1]/sin7x

    Can anyone explain how I can start this problem? I am clueless!
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    Re: find limit as x approach 0 : [cos(2x) -1]/sin7x

    You can use l'Hopital's rule.
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    Re: find limit as x approach 0 : [cos(2x) -1]/sin7x

    Quote Originally Posted by NeoSonata View Post
    find limit as x approach 0 : [cos(2x) -1]/sin7x

    Can anyone explain how I can start this problem? I am clueless!
    Hi NeoSonata,

    Since both the numerator and denominator tends to zero as x tends to zero, this is an indeterminate form. I suggest you use the L'H˘pital's rule.
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    Re: find limit as x approach 0 : [cos(2x) -1]/sin7x

    [(-2sin2x)(sin7x)-(7cos7x)(cos2x-1)]/(sin7x)^2

    Is this how it works? I know how to use product and quotient rule but I am unfamiliar with dealing with "sin7x", "cos2x-1", anything similar to that.
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    Re: find limit as x approach 0 : [cos(2x) -1]/sin7x

    That's not what L'Hopitals rule says, you have to take the derivative of the numerator and the derivative of the denominator, not of the fraction. Try it again.
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    Re: find limit as x approach 0 : [cos(2x) -1]/sin7x

    derivative of the numerator and the derivative of the denominator :

    -2sin2x / 7cos7x
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    Re: find limit as x approach 0 : [cos(2x) -1]/sin7x

    Correct. Can you determine the limit now? ...
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    Re: find limit as x approach 0 : [cos(2x) -1]/sin7x

    Quote Originally Posted by Sudharaka View Post
    Hi NeoSonata,

    Since both the numerator and denominator tends to zero as x tends to zero, this is an indeterminate form. I suggest you use the L'H˘pital's rule.
    The power series for cos(2x) and sin(7x) can also be substituted, the result simplified and then the limit easily calculated.
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    Re: find limit as x approach 0 : [cos(2x) -1]/sin7x

    Quote Originally Posted by NeoSonata View Post
    derivative of the numerator and the derivative of the denominator :

    -2sin2x / 7cos7x
    Yes that is correct. I think you have to understand the L'H˘pital's rule. So I will briefly explain. Suppose you have to evaluate a limit of the form, \lim_{x\rightarrow c}\frac{f(x)}{g(x)}=\frac{0}{0}\mbox{ or }\lim_{x\rightarrow c}\frac{f(x)}{g(x)}=\frac{\pm\infty}{\pm\infty} where f(x) and g(x) are two differentiable functions and c\in\Re\mbox{ or }c=\pm\infty

    Then the L'H˘pital's rule says that, \lim_{x\rightarrow c}\frac{f(x)}{g(x)}=\lim_{x\rightarrow c}\frac{f'(x)}{g'(x)}

    So in this case you have to evaluate, \lim_{x\rightarrow 0}\frac{\cos 2x-1}{\sin7x}=\lim_{x\rightarrow 0}\frac{-2\sin 2x}{7\cos 7x}=0

    A more detailed (and easily understandable) description about the L'H˘pital's rule could be found at, Pauls Online Notes : Calculus I - L'Hospital's Rule and Indeterminate Forms
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    Re: find limit as x approach 0 : [cos(2x) -1]/sin7x

    I would consider both "L'Hopital's rule" and "power series" to be 'overkill' for a problem like this. Write the fraction as
    \frac{cos(2x)- 1}{sin(7x)}= \left(\frac{2}{7}\right)\left(-\frac{1- cos(2x)}{2x}\right)\left(\frac{7x}{sin(7x)}\right)

    And then use \lim_{u\to 0}\frac{sin(u)}{u}= 1 and \lim_{v\to 0}\frac{1- cos(v)}{v}= 0
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    Re: find limit as x approach 0 : [cos(2x) -1]/sin7x

    Hello, NeoSonata!

    From the "shape" of the problem, I suspect these theorems are suggested:

    . . \lim_{\theta\to0}\frac{\cos\theta - 1}{\theta} \;=\;0   \qquad\quad \lim_{\theta\to0}\frac{\sin\theta}{\theta} \;=\;1



    \lim_{x\to0}\frac{\cos(2x) -1}{\sin(7x)}

    We have: . \frac{\cos(2x)-1}{1}\cdot \frac{1}{\sin(7x)}


    Multiply the first fraction by \tfrac{2x}{2x}, the second by \tfrac{7x}{7x}

    . . \displaystyle \frac{2x}{2x}\cdot\frac{\cos(2x)-1}{1}\cdot\frac{7x}{7x}\cdot\frac{1}{\sin(7x)} \;=\;\frac{2}{7}\cdot\frac{\cos(2x)-1}{2x}\cdot\frac{7x}{\sin(7x)}


    \text{When }x\to0,\,\text{ both }2x\to 0 \text{ and }7x\to0.

    \text{We have: }\:\lim_{x\to0}\left[\frac{2}{7}\cdot\frac{\cos(2x)-1}{2x}\cdot\frac{7x}{\sin(7x)}\right]

    . . . . . =\;\frac{2}{7}\left[\lim_{2x\to0}\frac{\cos(2x)-1}{2x}\right]\,\left[\lim_{7x\to0}\frac{7x}{\sin(7x)}\right]

    . . . . . =\;\frac{2}{7}\cdot 0 \cdot 1 \;\;=\;\;0


    Edit: Ah, HallsofIvy beat me to it! . . . *sigh*
    Last edited by Soroban; July 14th 2011 at 06:52 AM.
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