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  1. #1
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    limits

    lim ( 1 + 3/x) ^(.2x)
    x->infinity

    i have no idea how to solve this limit, if someone could point me in the right direction, that would be great. thanks!
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  2. #2
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    Put x\mapsto\frac1x and the limit becomes \lim_{x\to0}\left( 1+3x \right)^{2/x}, hence

    \left( 1+3x \right)^{2/x}=\exp \left( \frac{2}{x}\cdot \ln (1+3x) \right)=\exp \left( 6\cdot \frac{\ln (1+3x)}{3x} \right)\to e^{6} as x\to0.
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    thanks! okay this makes sense to me, but why did you have to substitute 1/x for x?
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  4. #4
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    Hello,
    Quote Originally Posted by sarabolha View Post
    lim ( 1 + 3/x) ^(.2x)
    x->infinity

    i have no idea how to solve this limit, if someone could point me in the right direction, that would be great. thanks!
    You can note that :
    \lim_{x \to \infty} \left(1+\frac ax\right)^x=e^a (it is provable by a simple substitution and using \left(1+\frac 1x\right)^x \to e

    Now, a^{bc}=(a^b)^c

    So \left(1+\frac 3x\right)^{0.2 \cdot x}=\left(\left(1+\tfrac 3x\right)^x\right)^{0.2} \to (e^3)^{0.2}=e^{0.6}
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by sarabolha View Post
    lim ( 1 + 3/x) ^(.2x)
    x->infinity

    i have no idea how to solve this limit, if someone could point me in the right direction, that would be great. thanks!
    Exactly as the others did except with a little more working

    \lim_{x\to\infty}\left(1+\frac{a}{x}\right)^{bx}

    If we let x=\frac{1}{z} this limit becomes

    \lim_{z\to{0}}\left(1+az\right)^{\frac{b}{z}}

    Now assuming the limit exists and equals L we have that

    \ln\left(L\right)=\lim_{z\to{0}}\ln\left(\left(1+a  z\right)^{\frac{b}{z}}\right)=b\lim_{z\to{0}}\frac  {\ln(1+az)}{z}

    Now remember what we know about derivatives f'(c)=\lim_{x\to{c}}\frac{f(x)-f(c)}{x-c}

    So in our integral we have that c=0 and f(x)=\ln(1+ax)\Rigtharrow{f(0)=\ln(1+0)=0

    So we have that

    \lim_{z\to{0}}\frac{\ln(1+az)}{z}=\lim_{z\to{0}}\f  rac{\ln(1+az)-\ln(1+a(0))}{z-0}=\bigg[\ln(1+az)\bigg]'\bigg|_{z=0}=\frac{a}{1+a(0)}=a

    \therefore\ln\left(L\right)=b\lim_{z\to{0}}\frac{\  ln(1+az)}{z}=ba\Rightarrow{L=e^{ab}}

    Just like the others said
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  6. #6
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    Quote Originally Posted by sarabolha View Post

    thanks! okay this makes sense to me, but why did you have to substitute 1/x for x?
    It's like putting z=\frac1x, there's no difference between \lim_{x\to0}x and \lim_{z\to0}z, both achieve to the same value.
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