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Math Help - Evaluate limit?

  1. #1
    Super Member fardeen_gen's Avatar
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    Evaluate limit?

    Evaluate:

    \lim_{x\rightarrow 0} \frac{27^x - 9^x - 3^x + 1}{\sqrt{2} - \sqrt{1 + \cos x}}
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  2. #2
    MHF Contributor Amer's Avatar
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    Jordan
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    Quote Originally Posted by fardeen_gen View Post
    Evaluate:

    \lim_{x\rightarrow 0} \frac{27^x - 9^x - 3^x + 1}{\sqrt{2} - \sqrt{1 + \cos x}}

    \lim_{x\rightarrow 0 } \frac{27^x-9^x-3^x+1}{\sqrt{2}-\sqrt{1+cosx}}=\frac{0}{0}=lop=

    <br />
lim_{x\rightarrow 0 } \frac{2(ln(27)27^x-ln(9)9^x-ln(3)3^x)(\sqrt{cosx+1})}{-sinx}=\frac{0}{0}=lop


    lim_{n\rightarrow 0 } \frac{2\sqrt{2}((ln27)^{2}27^x-(ln9)^{2}9^x-(ln3)^{2}3^x}{-cosx}=


    \frac{-2\sqrt{2}((3ln(3))^2-(2ln(3))^2-(ln(3))^2)}{1}=-2\sqrt{2}(5(ln(3))^2)
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  3. #3
    Super Member fardeen_gen's Avatar
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    The answer is
    Spoiler:
    8\sqrt{2}(\ln 3)^2
    . Where did we go wrong?
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  4. #4
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Amer View Post
    \lim_{x\rightarrow 0 } \frac{27^x-9^x-3^x+1}{\sqrt{2}-\sqrt{1+cosx}}=\frac{0}{0}=lop=

    <br />
lim_{x\rightarrow 0 } \frac{2(ln(27)27^x-ln(9)9^x-ln(3)3^x)(\sqrt{cosx+1})}{{\color{red}-}(-sinx)}=\frac{0}{0}=lop since cosx derivative is -sinx and you have negative sign so sinx should be instead of -sinx


    lim_{n\rightarrow 0 } \frac{2\sqrt{2}((ln27)^{2}27^x-(ln9)^{2}9^x-(ln3)^{2}3^x}{cosx}=


    \frac{2\sqrt{2}((3ln(3))^2-(2ln(3))^2-(ln(3))^2)}{1}=2\sqrt{2}({\color{red}5}(ln(3))^2)


    \frac{2\sqrt{2}((3ln(3))^2-(2ln(3))^2-(ln(3))^2)}{1}= (2\sqrt{2}((ln(3))^2(9-4-1))=2\sqrt{2}(4(ln(3))^2)
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