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Math Help - Limits - Evaluate without L'Hospital's rule or Series expansion?

  1. #1
    Super Member fardeen_gen's Avatar
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    Limits - Evaluate without L'Hospital's rule or Series expansion?

    Evaluate without L'Hospital's rule or Series Expansion:

    \lim_{x\rightarrow 4} \frac{(\cos \alpha)^x - (\sin \alpha)^x - \cos 2\alpha}{x - 4}
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  2. #2
    Member Ruun's Avatar
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    cos^4 (\alpha) - sin^4 (\alpha) =(cos^2(\alpha) + sin^2(\alpha))(cos^2(\alpha)-sin^2(\alpha))=(cos^2(\alpha)-sin^2(\alpha))

    As cos(2\alpha)=cos^2(\alpha)-sin^2(\alpha) the numerator is zero, and the denominator tends to zero... In this case the limit will be zero. But as x is not four, but tends to four, I'm not sure if the above trig. expressions can be used.
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  3. #3
    Moo
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    Hello,
    Quote Originally Posted by Ruun View Post
    cos^4 (\alpha) - sin^4 (\alpha) =(cos^2(\alpha) + sin^2(\alpha))(cos^2(\alpha)-sin^2(\alpha))=(cos^2(\alpha)-sin^2(\alpha))
    That's correct til here.

    As cos(2\alpha)=cos^2(\alpha)-sin^2(\alpha) the numerator is zero, and the denominator tends to zero... In this case the limit will be zero. But as x is not four, but tends to four, I'm not sure if the above trig. expressions can be used.
    0/0 is an undetermined limit !!!!!!


    Anyway, then you can notice that \cos^4\alpha-\sin^4\alpha=\cos^2\alpha-\sin^2\alpha=\cos(2\alpha)

    So if you let f(x)=\cos^x\alpha-\sin^x\alpha, your limit is actually :
    \lim_{x\to 4} \frac{f(x)-f(4)}{x-4}, which is (the difference quotient of the function f) equal to f'(4)
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  4. #4
    Senior Member pankaj's Avatar
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    \lim_{x\rightarrow 4} \frac{(\cos \alpha)^x - (\sin \alpha)^x - \cos 2\alpha}{x - 4}

    =\lim_{x\rightarrow 4} \frac{(\cos \alpha)^x - (\sin \alpha)^x - \cos^4\alpha+\sin^4\alpha}{x - 4}


    =\lim_{x\rightarrow 4} \frac{(\cos \alpha)^x -\cos^4\alpha- ((\sin \alpha)^x -\sin^4\alpha)}{x - 4}


    =\lim_{x\rightarrow 4} \frac{\cos^4\alpha((\cos \alpha)^{x-4} -1)-\sin^4\alpha ((\sin \alpha)^{x-4} -1)}{x - 4}


    =\cos^4\alpha.ln(\cos\alpha)-\sin^4\alpha.ln(\sin\alpha)
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