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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?)

    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?

    P.S. I tested the values on a calculator.
    Do not double post.

    You can be banded for that.
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  3. #3
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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?

    P.S. I tested the values on a calculator.

    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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