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Math Help - need a little limit help action

  1. #1
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    need a little limit help action

    It can be shown that .
    Use this limit to evaluate the limits below.

    (a) =

    (b) =

    need some direction/help...thanks
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    For the first one, define u=\frac1x, so

    \lim_{x\to0}(1+x)^{1/x}=\lim_{u\to\infty}\left(1+\frac1u\right)^u=e.

    For the second one, you can take it more generally:

    \forall x\in\mathbb R,\,\lim_{n\to\infty}\left(1+\frac xn\right)^n=e^x.
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  3. #3
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    thanks for the help, i know these are right, however when i try to submit them, it tells me they are wrong...do you have any idea why...

    thanks
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  4. #4
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by mer1988 View Post
    It can be shown that .
    Use this limit to evaluate the limits below.

    (a) =

    (b) =

    need some direction/help...thanks
    the previous solution was also the first thing that came into my mind, yet your teacher didn't want it that way.. i haev another approach..
    we know that f(x)=e^{\ln {f(x)}}\, thus if we let f(x)= (1+x)^{\frac{1}{x}}\,, then we have
    e^{\ln {(1+x)^{\frac{1}{x}}}} = e^{\frac{1}{x} \ln {(1+x)}}, and take the limit as x \rightarrow 0^+ so it becomes
    \lim_{x \rightarrow 0^+}e^{\frac{1}{x} \ln {(1+x)}} = e^{\lim_{x \rightarrow 0^+}{\frac{1}{x} \ln {(1+x)}}} = e^{\lim_{x \rightarrow 0^+}{\frac{\ln {(1+x)}}{x}}}
    and the limit part is in the case of \frac{0}{0}, you may use L'hospital's Rule, so the limit part equals 1 and so e^1 = e..

    you can do the same thing for 2..
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