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Math Help - L'Hopital/limits question

  1. #1
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    L'Hopital/limits question

    Here's the question:
    lim(x-->infinity) x/((1+x^2)^(1/2))

    One approach (that Wolfram Alpha used):
    =square root of lim x^2/(1+x^2) = 1
    (using L'hopital)

    Another approach:
    Using L'hopital: =((1+x^2)^(1/2))/x
    which is our original question upside down.
    Does this prove the limit is 1.

    ie. if limit a/b = limit b/a does this mean the limit must equal 1?

    Thanks
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  2. #2
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    \[\dfrac{x}{{\sqrt {1 + x^2 } }} = \dfrac{1}{{\sqrt {\frac{1}{{x^2 }} + 1} }}
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    No. Try to do the same to f(x)=x/e^x (for your first question!)
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  4. #4
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    Quote Originally Posted by durrrrrrrr View Post
    ie. if limit a/b = limit b/a does this mean the limit must equal 1?
    If both limits exist

    limit a/b = 1/(limit b/a)

    So limit a/b could also be -1.
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Plato View Post
    \[\dfrac{x}{{\sqrt {1 + x^2 } }} = \dfrac{1}{{\sqrt {\frac{1}{{x^2 }} + 1} }}
    It should be:

    f(x)=\dfrac{x}{{\sqrt {1 + x^2 } }} =\begin{Bmatrix} \dfrac{1}{{\sqrt {\frac{1}{{x^2 }} + 1} }} & \mbox{ if }& x> 0\\-\dfrac{1}{{\sqrt {\frac{1}{{x^2 }} + 1} }} & \mbox{if}& x<0\end{matrix}

    So,

    \displaystyle\lim_{x \to{+}\infty}{f(x)}=1,\quad \displaystyle\lim_{x \to{-}\infty}{f(x)}=-1

    and does not exist

    \displaystyle\lim_{x \to{}\infty}{f(x)}




    Fernando Revilla
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  6. #6
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    \displaystyle\lim_{x\to\infty}\frac{x}{\sqrt{x^2+1  }}\Rightarrow\lim_{x\to\infty}\frac{x}{\sqrt{x^2}}  \Rightarrow\lim_{x\to\infty}\frac{x}{\pm x}

    \displaystyle\lim_{x\to\infty}\frac{x}{x}=1 \ \mbox{and} \ \lim_{x\to\infty}\frac{x}{-x}=-1
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  7. #7
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    Because the OP is \displaystyle \lim _{x \to \infty } \frac{x}{{\sqrt {1 + x^2 } }}, we can assume that x>0
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  8. #8
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Plato View Post
    Because the OP is \displaystyle \lim _{x \to \infty } \frac{x}{{\sqrt {1 + x^2 } }}, we can assume that x>0
    No, you can't assume that. The formula that defines f has sense on \mathbb{R} .When a function is given by a formula an the domain is not specified it is an universal convenious to apply the rule of the maximum domain i.e, D(f)=\mathbb{R}, and topologically \infty\neq +\infty .


    Fernando Revilla
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  9. #9
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    Quote Originally Posted by Plato View Post
    Because the OP is \displaystyle \lim _{x \to \infty } \frac{x}{{\sqrt {1 + x^2 } }}, we can assume that x>0
    If we want to assume x>0, then shouldn't the limit be written as \displaystyle\lim_{x\to\infty^+}\frac{x}{\sqrt{x^2  +1}}\mbox{?}
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  10. #10
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by dwsmith View Post
    If we want to assume x>0, then shouldn't the limit be written as \displaystyle\lim_{x\to\infty^+}\frac{x}{\sqrt{x^2  +1}}\mbox{?}

    Right.


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  11. #11
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    Quote Originally Posted by FernandoRevilla View Post
    Hence, the limit DNE since the (+) is infinity and (-) is negative infinity.
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  12. #12
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by dwsmith View Post
    Hence, the limit DNE since the (+) is infinity and (-) is negative infinity.
    Right.

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  13. #13
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    Quote Originally Posted by FernandoRevilla View Post
    No, you can't assume that. The function f is defined on \mathbb{R} .When a function is given by a formula an the domain is not specified it is an universal convenious to apply the rule of the maximum domain i.e, D(f)=\mathbb{R}, and topologicaly \infty\neq +\infty .
    With all due respect, I think you have no idea about how this material is currently being taught. What you have written above is not consistent with any standard calculus text book in use today. I have taught calculus classes since 1964.
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  14. #14
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    Does not the notation

    \displaystyle\lim_{x \to a^{+}}

    denote the limit as "x" approaches "a" from the right?

    In which case

    \displaystyle\lim_{x \to \infty^{+}}

    requires x<0 and approaching -\infty
    Last edited by Archie Meade; January 9th 2011 at 02:35 AM. Reason: clarification
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  15. #15
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    Quote Originally Posted by Plato View Post
    With all due respect, I think you have no idea about how this material is currently being taught. What you have written above is not consistent with any standard calculus text book in use today. I have taught calculus classes since 1964.
    Out of curiousity you are saying when infinity is printed in a limit in a Calculus text book, the book is implying the positive values only?

    Also, my Calc text book isn't written in such a manner. If there isn't a + or -, the it is the overall limit. My book is very good about specifying direction.
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