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Math Help - L'Hospital's Rule

  1. #1
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    L'Hospital's Rule

    lim -> infinite (3x+4)/(ln(4+4e^x))
    lim -> 0 (2/x^4)-(5/x^2)
    lim -> infinite (1+(10/x))^(x/4)
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Calc1stu View Post
    lim -> infinite (3x+4)/(ln(4+4e^x))
    lim -> 0 (2/x^4)-(5/x^2)
    lim -> infinite (1+(10/x))^(x/4)
    Come on...you don't need L'hopitals for any of these. Just think about this

    As x gets incredibely huge 4+4e^x\to{4e^x}

    So \ln(4+4e^x)\to\ln(4)+x

    So \frac{3x+4}{\ln(4+4e^x)}\to\frac{3x+4}{\ln(4)+x}\t  o{3}

    For the secone one combining the fractions give \frac{2-5x^2}{x^4}} now when you put 0 in there it should be pretty obvious what the answer is


    For the third one Im sure you know that

    \lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x=e

    Well in your limit if we let \frac{10}{x}=u we have that

    \lim_{x\to\infty}\left(1+\frac{10}{x}\right)^{\fra  c{x}{4}}=\lim_{u\to{0}}\left(1+u\right)^{\frac{10}  {4u}}

    Now we can see that this is equivalent to

    \left[\lim_{u\to{0}}\left(1+u\right)^{\frac{1}{u}}\right]^{\frac{10}{4}}

    On the inner integral letting \frac{1}{x}=u again we get

    \left[\lim_{x\to\infty}\left(1+\frac{1}{x}\right)\right]^{\frac{10}{4}}=e^{\frac{10}{4}}
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