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Math Help - Sepecial limit

  1. #1
    Super Member dhiab's Avatar
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    Sepecial limit

    Calculate :
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by dhiab View Post
    Calculate :

    \sqrt[4]{x^4+1}=\sqrt[4]{x^4(1+\frac{1}{x^4})}=x\sqrt[4]{1+\frac{1}{x^4}}

    since \sqrt[4]{1+\frac{1}{x^4}}\to{1} as x\to{+\infty}, by substitution we have +\infty(1)-(+\infty)=huh?
    Last edited by VonNemo19; June 8th 2009 at 03:07 PM.
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  3. #3
    Math Engineering Student
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    if you're trying to say that \infty-\infty=0, your solution is wrong.
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Krizalid View Post
    if you're trying to say that \infty-\infty=0, your solution is wrong.

    Well, I admit that I don't have a good grasp on limits of this nature, but I do know that

    \lim_{x\to{c}}f(x)*g(x)=[\lim_{x\to{c}}f(x)][\lim_{x\to{c}}g(x)]

    and that the limit of a sum is the sum of the limits.

    I know I used good math here. The only part that I was unsure of was subtraccting infinity from infinity.

    Could you please enlighten me with an explaination?
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  5. #5
    Super Member Random Variable's Avatar
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    Multlpyling and dividing by the conjugate doesn't seem to help.

     \lim_{x \to \infty} (\sqrt[4]{x^{4}+1} - x) \frac {\sqrt[4]{x^{4}+1} + x}{\sqrt[4]{x^{4}+1} + x}

    = \lim_{x \to \infty} \frac {\sqrt{x^{4}+1} - x^{2}}{\sqrt[4]{x^{4}+1} + x}

    but then the limit is indeterminate  \big(\frac {0}{0}\big) , which can been seen if you divide the top and bottom by x^{2}

    I guess you could then use L'Hospital's rule, but it gets messy.
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  6. #6
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Random Variable View Post
    Multlpyling and dividing by the conjugate doesn't seem to help.

     \lim_{x \to \infty} (\sqrt[4]{x^{4}+1} - x) \frac {\sqrt[4]{x^{4}+1} + x}{\sqrt[4]{x^{4}+1} + x}

    = \lim_{x \to \infty} \frac {\sqrt{x^{4}+1} - x^{2}}{\sqrt[4]{x^{4}+1} + x}

    but then the limit is indeterminate  \big(\frac {0}{0}\big) , which can been seen if you divide the top and bottom by x^{2}

    I guess you could then use L'Hospital's rule, but it gets messy.
    Applying L'Hopital

    = \lim_{x \to \infty} \frac {\sqrt{x^{4}+1} - x^{2}}{\sqrt[4]{x^{4}+1} + x}

    =\frac{x^3(x^4+1)^{-3/4}-2x}{x^3(x^4+1)^{-3/4}+1}


    =\frac{(x^4+1)^{-3/4}-2x}{(x^4+1)^{-3/4}+1}

    =\frac{(x^4+1)^{-3/4}}{(x^4+1)^{-3/4}+1}\to{1}

    and

    let u=(x^4+1)

    =\frac{(u)^{-3/4}}{(u)^{-3/4}+1}\to{1}

    someone that knows what they're doing should finish this. I suck
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  7. #7
    Super Member Random Variable's Avatar
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    I feel like a total idiot for not seeing an easy way to do this. The limit is obviously 0, though.
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  8. #8
    Member javax's Avatar
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    here's my effort
    \lim_{x\to\infty}{\sqrt[4]{x^4+1}-x}
    = \lim_{x\to\infty}{\sqrt[4]{x^4+1}-x\left(\frac{\sqrt[4]{(x^4+1)^3}+\sqrt[4]{x^4+1}*x^2+\sqrt{x^4+1}*x+x^3}{\sqrt[4]{(x^4+1)^3}+\sqrt[4]{x^4+1}*x^2+\sqrt{x^4+1}*x+x^3}\right)}
    =\lim_{x\to\infty}\frac{1}{x^3\sqrt[4]{(1+\frac{1}{x^4})^3}+x^3\sqrt[4]{1+\frac{1}{x^4}}+x^4\sqrt[4]{1+\frac{1}{x^2}}+x^3}

    now divide by x^4 and puting the limit it will be

    \frac{0}{0+0+1+0} = 0

    correct me if i made a mistake.
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  9. #9
    No one in Particular VonNemo19's Avatar
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    I'm just gonna stay on my level for now. i thank you all.
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  10. #10
    Flow Master
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    Quote Originally Posted by dhiab View Post
    Calculate :
    Substitute x = \frac{1}{t} and consider the limit t \rightarrow 0.

    If this had been posted in the Calculus subforum you could now easily apply l'Hopital's rule.

    However, without using l'Hopital's Rule, there's slightly (but not a lot) more work ..... multiply numerator and denominator by the conjugate surd (twice) and then cancel a common factor of t.

    Either way, you get 0 as the answer.
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  11. #11
    Grand Panjandrum
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    Quote Originally Posted by dhiab View Post
    Calculate :
    (x^4+1)^{1/4}-x=x(1+1/x^4)^{1/4}-x=x\left(1+\frac{1}{4x^4}+O(x^{-8})\right)-x=\frac{1}{4x^3}+O(x^{-7})

    Hence:

     <br />
\lim_{x\to \infty}\left((x^4+1)^{1/4}-x\right)=0<br />

    Supporting evidence for this can be found by brute computation.

    CB
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  12. #12
    No one in Particular VonNemo19's Avatar
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    I would just like to say that I appreciate how helpful Captain Black and Mr.fantastic have been. I was racking my brain for hours over this one. I knew that there were easier ways.
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  13. #13
    Grand Panjandrum
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    Quote Originally Posted by CaptainBlack View Post
    (x^4+1)^{1/4}-x=x(1+1/x^4)^{1/4}-x=x\left(1+\frac{1}{4x^4}+O(x^{-8})\right)-x=\frac{1}{4x^3}+O(x^{-7})

    Hence:

     <br />
\lim_{x\to \infty}\left((x^4+1)^{1/4}-x\right)=0<br />

    Supporting evidence for this can be found by brute computation.

    CB
    If you are not happy with big-O notation you can use the remainder for the Taylor expansion instead.

    CB
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  14. #14
    Super Member dhiab's Avatar
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    Another resolution

    Hello Here is another resolution :
    On using this identity :
    and put :

    they acquire :
    they will have as zero
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