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Math Help - Limit problem (what's going wrong?)

  1. #1
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    Limit problem (what's going wrong?)

    {\lim _{x \to - \infty }}{\frac{4x}{\sqrt{x^{2}+4x}+x}}}

    This gives - \infty, which is correct.

    Dividing both the numerator and the denominator by x, we have

    {\lim _{x \to - \infty }}{\frac{4}{\sqrt{1+\frac{4}{x}}+1}}}

    now this gives me 2.

    What's wrong?

    P.S. I tested the values on a calculator.
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  2. #2
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    Re: Limit problem (what's going wrong?)

    Why do you feel it is wrong and what do you think is wrong? have confidence in yourself.
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  3. #3
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    Re: Limit problem (what's going wrong?)

    I mean they even become two different functions after the division. Their behaviors differ when x<-4.
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  4. #4
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    Re: Limit problem (what's going wrong?)

    Quote Originally Posted by HeilKing View Post
    {\lim _{x \to - \infty }}{\frac{4x}{\sqrt{x^{2}+4x}+x}}}
    This gives - \infty, which is correct.
    Dividing both the numerator and the denominator by x, we have
    {\lim _{x \to - \infty }}{\frac{4}{\sqrt{1+\frac{4}{x}}+1}}}
    now this gives me 2.
    What's wrong?.
    Put your calculator away. Use your brain.

    If x<0 then {\frac{4x}{\sqrt{x^{2}+4x}+x}}<0. So what is the limit?
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  5. #5
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    Re: Limit problem (what's going wrong?)

    I think that the first answer is incorrect. I think that the value of the limit is +\infty, not -\infty.

    If x(<-4) is negative then \sqrt{x^{2}+4x} is less than x in magnitude which implies that the denominator is negative and the fraction positive.

    There is a presumption in the notation that \sqrt{x^{2}+4x} is a positive quantity. What is happening when the x is cancelled is as if a -\infty is coming out from the square root in effect making it negative. (That is, \sqrt{x^2+4x} tends to -\infty rather than +\infty as it should).

    A way round the problem is to replace x by -x and take the limit as x \rightarrow +\infty.
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  6. #6
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    Re: Limit problem (what's going wrong?)

    In the first form, if you are "taking x going to -\infty", the numerator goes to -\infty goes to -\infty while the denominator goes to \infty- \infty, an indeterminant form. Notice that if x= -10000, then \frac{4x}{\sqrt{x^2+ 4x}+ x}= \frac{-40000}{\sqrt{9999}- 10000}= \frac{-40000}{9999.5}= -4.0002 which is NOT going to " -\infty".
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  7. #7
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    Re: Limit problem (what's going wrong?)

    Hi,
    BobP is correct. The limit is +\infty.
    The problem is some subtle elementary algebra. For x < 0, \sqrt{x^2}=-x. So for x < 0,
    {\sqrt{x^2+4x}\over x}={\sqrt{x^2+4x}\over-\sqrt{x^2}}=-{\sqrt{{x^2+4x\over x^2}}=-\sqrt{1+{4\over x}}
    Thus for x < 0, {4x\over\sqrt{x^2+4x}+x}={4\over-\sqrt{1+{4\over x}}+1
    Since x < 0, the denominator is positive and approaches 0 as x goes to infinity. So the original limit is plus infinity.
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  8. #8
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    Re: Limit problem (what's going wrong?)

    Quote Originally Posted by HallsofIvy View Post
    Notice that if x= -10000, then \frac{4x}{\sqrt{x^2+ 4x}+ x}= \frac{-40000}{\sqrt{9999}- 10000}= \frac{-40000}{9999.5}= -4.0002 which is NOT going to " -\infty".
    A correction here, maybe a typo, but it changes a lot.
    \frac{4x}{\sqrt{x^2+ 4x}+ x}= \frac{-40000}{9999 - 10000}= \frac{-40000}{-1}= 40000

    Let's try this for a larger x, say x = -10^9. Then
    \frac{4x}{\sqrt{x^2+ 4x}+ x} = \frac{-4 \times 10^9}{ 10^9 -10^9 }
    The denominator here isn't really 0, just that the number from the square root is very close to 10^9, and smaller than the 10^9 from the x term in the denominator...it also makes the denominator negative. So the expression is very much larger than +10^9. Sounds like it's headed for infinity to me.

    -Dan
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  9. #9
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    Re: Limit problem (what's going wrong?)

    Since x is going to -\infty, write x as -|x|. The given limit is
    \lim_{|x|\to\infty} \frac{-4|x|}{\sqrt{|x|^2 -4|x|} -|x|}
    = \lim_{|x|\to\infty} \frac{-4}{\sqrt{1 -\frac{4}{|x|}} -1}
    = \lim_{\epsilon\downarrow 0} \frac{4}{1 - \sqrt{1 -\epsilon} } = +\infty
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  10. #10
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    Re: Limit problem (what's going wrong?)

    first it can be written as

    \[\mathop {\lim }\limits_{x \to  - \infty } \frac{{4x}}{{\sqrt {{x^2} + 4x}  + x}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{4x}}{{\sqrt {{x^2} + 4x}  + x}}*(\sqrt {{x^2} + 4x}  - x)/(\sqrt {{x^2} + 4x}  - x) = \mathop {\lim }\limits_{x \to  - \infty } \frac{{4x}}{{4x}}*(\sqrt {{x^2} + 4x}  - x) = \mathop {\lim }\limits_{x \to  - \infty } \sqrt {{x^2} + 4x}  - x\]


    now you cannot take x out since it is a negative number

    this question can be done in the following way

    since x is negative , convert it to a positive number by substituting y= -x or x= -y

    therefore functon becomes \[\mathop {\lim }\limits_{x \to  - \infty } (\sqrt {{x^2} + 4x}  - x) = \mathop {\lim }\limits_{y \to \infty } \sqrt {{y^2} - 4y}  + y = \mathop {\lim }\limits_{y \to \infty } y(\sqrt {1 - 4/y}  + 1) = \mathop {\lim }\limits_{y \to \infty } y(1 + 1) = \mathop {\lim }\limits_{y \to \infty } 2y = \infty \]

    thus answer is infinity not minus infinity......
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