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Thread: Limit problem (what's going wrong?)

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

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

    This gives $\displaystyle - \infty$, which is correct.

    Dividing both the numerator and the denominator by $\displaystyle x$, we have

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

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

    This gives $\displaystyle - \infty$, which is correct.

    Dividing both the numerator and the denominator by $\displaystyle x$, we have

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

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

    This gives $\displaystyle - \infty$, which is correct.

    Dividing both the numerator and the denominator by $\displaystyle x$, we have

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

    now this gives me 2.

    What's wrong?

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

    first it can be written as

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


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

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