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Math Help - Need Help with 3 Limit Problems

  1. #1
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    Need Help with 2 Limit Problems

    Hi, I need help with 2 limits problems. I can't figure them out. Please help me.

    i know the first one just the other 2!!! thanks!!1
    Last edited by greenbee; September 29th 2009 at 10:16 AM.
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by greenbee View Post
    Hi, I need help with 2 limits problems. I can't figure them out. Please help me.

    i know the first one just the other 2!!! thanks!!1

    \lim_{x\rightarrow \infty} \frac{\cos x }{e^{4x}} dx =0

    since maximum value of cos x is 1

    \lim_{x\rightarrow \infty} \frac{x+2}{\sqrt{64x^2 +1}} dx

    divide the denominator and the numerator by x so

    \lim_{x\rightarrow \infty} \frac{1 +\frac{2}{x}}{\sqrt{64 + \frac{1}{x^2}}}= \frac{1}{\sqrt{64}}


    that's it
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  3. #3
    MHF Contributor Amer's Avatar
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    final question I will solve it without graph like this


    \lim_{x\rightarrow \infty} \left( 1- \frac{3}{x} \right)^x

    let \lim_{x\rightarrow \infty} \left( 1- \frac{3}{x} \right)^x = u

    so

    \ln u = \ln \lim_{x\rightarrow \infty} \left( 1- \frac{3}{x} \right)^x

    \ln u = \lim_{x\rightarrow \infty}\ln \left( 1- \frac{3}{x} \right)^x

    \lim_{x\rightarrow \infty} x \ln \left( 1- \frac{3}{x} \right)

    \lim_{x\rightarrow \infty} \frac{\ln \left( 1- \frac{3}{x} \right)}{\frac{1}{x}}

    when you sub infinity instead of x you will have 0/0 so use lopital rule

    \lim_{x\rightarrow \infty} \left<br />
(\dfrac{\dfrac{\frac{3}{x^2}}{1-\frac{3}{x}}}{\dfrac{-1}{x^2}}\right)

    \lim_{x\rightarrow \infty} \frac{\left(\frac{3}{x^2}\right)\left(\frac{x}{x-3}\right)}{\frac{-1}{x^2}}

    \lim_{x\rightarrow \infty} \frac{-3x}{x-3} = -3 now

     \ln u = -3 \Rightarrow u = e^{-3} \Rightarrow u =\frac{1}{e^3} = 0.0497870683679
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  4. #4
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    Thank you so much for helping me!!! *bows*
    *click on thank you button nonstop!*
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