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Math Help - Evaluate a limit

  1. #1
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    Evaluate a limit

    Hey everyone,

    Got this problem from a math packet. No idea how to evaluate this limit. I assume L'hopital's rule is involved.

    Evaluate a limit-codecogseqn-6-.gif


    No matter how hard I try, I am unable to manipulate this limit in such a way as to evaluate it.
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Evaluate a limit

    Have you tried taking the natural log, so that you get the form 0/0?
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  3. #3
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    Re: Evaluate a limit

    First obvious step is to factor out \frac{1}{2}. I would then factor out a^x so this becomes \frac{a}{2}(1+ (b/a)^x)^{1/x}.

    Now, do you know the limit definition of "e"?
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Evaluate a limit

    Quote Originally Posted by HallsofIvy View Post
    First obvious step is to factor out \frac{1}{2}. I would then factor out a^x so this becomes \frac{a}{2}(1+ (b/a)^x)^{1/x}.

    Now, do you know the limit definition of "e"?
    I don't think you can factor out \frac{1}{2}.
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  5. #5
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    Re: Evaluate a limit

    Quote Originally Posted by HallsofIvy View Post
    First obvious step is to factor out \frac{1}{2}. I would then factor out a^x so this becomes \frac{a}{2}(1+ (b/a)^x)^{1/x}
    Quote Originally Posted by MarkFL View Post
    I don't think you can factor out \frac{1}{2}.
    That is correct. It cannot be factored in that way.

    [URL="[QUOTE=HallsofIvy;794322]First obvious step is to factor out \frac{1}{2}. I would then factor out a^x so this becomes \frac{a}{2}(1+ (b/a)^x)^{1/x}. Now, do you know the limit definition of "e"?[/QUOTE]"

    Look at this link.
    http://www.wolframalpha.com/input/?i...9%2F%28x%29%29
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  6. #6
    Senior Member x3bnm's Avatar
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    Re: Evaluate a limit

    Quote Originally Posted by ShadowKnight8702 View Post
    Hey everyone,

    Got this problem from a math packet. No idea how to evaluate this limit. I assume L'hopital's rule is involved.

    Click image for larger version. 

Name:	CodeCogsEqn (6).gif 
Views:	8 
Size:	1.0 KB 
ID:	28945


    No matter how hard I try, I am unable to manipulate this limit in such a way as to evaluate it.
    \begin{align*}&\lim_{x\to 0}\left(\frac{a^x + b^x}{2}\right)^{\frac{1}{x}}\\=& \lim_{x\to 0}e^{\left(\ln\left(\frac{a^x + b^x}{2}\right)^{\frac{1}{x}}\right)}\\=& \lim_{x\to 0} e^{\left(\frac{1}{x}\ln{\left(a^x + b^x\right)} - \frac{1}{x}\ln{2}\right)}\\ &  \\=& \lim_{x\to 0} \frac{ \left( e^{\frac{1}{x}} \right) \left(e^{\ln{\left(a^x + b^x \right)}}\right)}{\left( e^{\frac{1}{x}}\right) \left(e^{\ln{2}}\right)}\\& \\=& \lim_{x\to 0} \frac{\left(a^x + b^x\right)}{2}\\=& \frac{(1+1)}{2} \\=& 1........\text{[Answer]}\end{align*}

    Hope it helps.
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  7. #7
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    Re: Evaluate a limit

    Quote Originally Posted by x3bnm View Post
    \begin{align*}&\lim_{x\to 0}\left(\frac{a^x + b^x}{2}\right)^{\frac{1}{x}}\\=& \lim_{x\to 0}e^{\left(\ln\left(\frac{a^x + b^x}{2}\right)^{\frac{1}{x}}\right)}\\=& \lim_{x\to 0} e^{\left(\frac{1}{x}\ln{\left(a^x + b^x\right)} - \frac{1}{x}\ln{2}\right)}\\ &  \\=& \lim_{x\to 0}\color{red} \frac{ \left( e^{\frac{1}{x}} \right) \left(e^{\ln{\left(a^x + b^x \right)}}\right)}{\left( e^{\frac{1}{x}}\right) \left(e^{\ln{2}}\right)}\\& \end{align*}
    @x3bnm;795132
    \left( e^{\frac{1}{x}} \right) \left(e^{\ln{\left(a^x + b^x \right)}}\right)\ne e^{\left(\frac{1}{x}\ln{\left(a^x + b^x\right)} \right)}

    We add exponents not multiply them.
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  8. #8
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    Re: Evaluate a limit

    This is a bit rough and ready, but I think that the result is okay.

    Suppose that b>a, then

    \left(\frac{a^{x}+b^{x}}{2}\right)^{1/x}=b\left(\frac{(a/b)^{x}+1}{2}\right)^{1/x}\rightarrow b, as x \rightarrow \infty,
    (since (a/b)^{x}\rightarrow 0, and the resulting (1/2)^{1/x}\rightarrow 1).

    Notice btw that if a=b then

    \left(\frac{a^{x}+b^{x}}{2}\right)^{1/x}=a, (or b).
    Thanks from ShadowKnight8702
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  9. #9
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    Re: Evaluate a limit

    Quote Originally Posted by BobP View Post
    This is a bit rough and ready, but I think that the result is okay.

    Suppose that b>a, then

    \left(\frac{a^{x}+b^{x}}{2}\right)^{1/x}=b\left(\frac{(a/b)^{x}+1}{2}\right)^{1/x}\rightarrow b, as x \rightarrow \infty,
    (since (a/b)^{x}\rightarrow 0, and the resulting (1/2)^{1/x}\rightarrow 1).

    Notice btw that if a=b then

    \left(\frac{a^{x}+b^{x}}{2}\right)^{1/x}=a, (or b).
    Please look at this page.

    NOTE (a/b)^{x}\not\rightarrow 0, you have (a/b)^{x}\rightarrow 1,

    On the wolfram page l'Hospitial's rule gives \exp\left(\frac{\ln(ab)}{2}\right)
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  10. #10
    Senior Member x3bnm's Avatar
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    Re: Evaluate a limit

    Quote Originally Posted by ShadowKnight8702 View Post
    Hey everyone,

    Got this problem from a math packet. No idea how to evaluate this limit. I assume L'hopital's rule is involved.

    Click image for larger version. 

Name:	CodeCogsEqn (6).gif 
Views:	8 
Size:	1.0 KB 
ID:	28945


    No matter how hard I try, I am unable to manipulate this limit in such a way as to evaluate it.
    \begin{align*}&\lim_{x\to 0} \left(\frac{(a^x + b^x)}{2}\right)^{\frac{1}{x}}\\=& \lim_{x\to 0} e^{\ln{\left(\frac{(a^x + b^x)}{2}\right)^{\frac{1}{x}}}}\\=& \lim_{x\to 0} e^{\frac{\ln{\left(a^x + b^x\right)}-\ln(2)}{x}}\end{align*}

    Now using I'Hopital's rule and property of limit this last line becomes:

    \begin{align*}&e^{\lim_{x\to 0} \frac{a^x \ln{(a)} + b^x \ln{(b)}}{a^x + b^x}}\\=& e^{\lim_{x\to 0} \frac{a^0 \ln{(a)} + b^0 \ln{(b)}}{a^0 + b^0}}\\=& e^{\left(\frac{\ln{(ab)}}{2}\right)}\\=& ab - e^2\text{.......[Answer]}\end{align*}
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  11. #11
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    Re: Evaluate a limit

    thanks x3bnm. The final step actually tursn the answer to sqrt(a)*sqrt(b) rather than ab-e^2, the process helped a lot.
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  12. #12
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    Re: Evaluate a limit

    Quote Originally Posted by x3bnm View Post
    \begin{align*}&\lim_{x\to 0} \left(\frac{(a^x + b^x)}{2}\right)^{\frac{1}{x}}\\=& \lim_{x\to 0} e^{\ln{\left(\frac{(a^x + b^x)}{2}\right)^{\frac{1}{x}}}}\\=& \lim_{x\to 0} e^{\frac{\ln{\left(a^x + b^x\right)}-\ln(2)}{x}}\end{align*}

    Now using I'Hopital's rule and property of limit this last line becomes:

    \color{red}\begin{align*}&e^{\lim_{x\to 0} \frac{a^x \ln{(a)} + b^x \ln{(b)}}{a^x + b^x}}\\=& e^{\lim_{x\to 0} \frac{a^0 \ln{(a)} + b^0 \ln{(b)}}{a^0 + b^0}}\\=& e^{\left(\frac{\ln{(ab)}}{2}\right)}\\=& ab - e^2\text{.......[Answer]}\end{align*} No indeed!
    \exp{\left(\frac{\ln{(ab)}}{2}\right)}=\exp\left( \ln{(\sqrt{ab}) \right)}=\sqrt{ab}

    @x3bnm: Please make sure that you understand the process before posting.
    Did you look at either of the Wolfram webpages links that I posted in replies #5 & #9?
    If not, why not?
    Last edited by Plato; August 21st 2013 at 03:48 PM.
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  13. #13
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    Re: Evaluate a limit

    Hi Plato

    Regarding your comment to my post #8, your post #9, somewhat sheepishly I have to admit that, (have just noticed that), I was calculating the wrong limit, x\rightarrow \infty rather than x\rightarrow 0.

    Apologies to all.
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