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Math Help - Limits going to negative infinity

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    Limits going to negative infinity

    \lim_{x \to -\infty } \frac{\sqrt{2x^2 + 1}}{3x - 5}\\\lim_{x \to -\infty } \frac{(\sqrt{2x^2 + 1}) \cdot \frac{1}{\sqrt{x^2}}}{(3x - 5) \cdot \frac{1}{x}}\\= \frac{\sqrt{2}}{3}

    But the solution says it's negative. Why is it negative? Is it just because the limit is going to negative infinity? So every time I take a limit going to negative infinity, should I just make my answer negative?
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    Re: Limits going to negative infinity

    Quote Originally Posted by PhizKid View Post
    \lim_{x \to -\infty } \frac{\sqrt{2x^2 + 1}}{3x - 5}\\\lim_{x \to -\infty } \frac{(\sqrt{2x^2 + 1}) \cdot \frac{1}{\sqrt{x^2}}}{(3x - 5) \cdot \frac{1}{x}}\\= \frac{\sqrt{2}}{3}

    But the solution says it's negative. Why is it negative? Is it just because the limit is going to negative infinity? So every time I take a limit going to negative infinity, should I just make my answer negative?
    \displaystyle \begin{align*} \sqrt{x^2} = |x| = \begin{cases} x \textrm{ if } \phantom{-}x \geq 0 \\ -x \textrm{ if } x < 0 \end{cases}  \end{align*}

    Since we are approaching \displaystyle \begin{align*} -\infty \end{align*}, will our x values be positive or negative?
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    Re: Limits going to negative infinity

    Well, looking at the function as a whole: we'd be taking very large negative values of x. So the denominator would become a very large negative and the numerator will always be positive. Is this the reason why? Or is there a process during taking the limit when you change signs somewhere?
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    Re: Limits going to negative infinity

    Quote Originally Posted by PhizKid View Post
    Well, looking at the function as a whole: we'd be taking very large negative values of x. So the denominator would become a very large negative and the numerator will always be positive. Is this the reason why? Or is there a process during taking the limit when you change signs somewhere?
    Please reread what I wrote. You have divided top and bottom by \displaystyle \begin{align*} \sqrt{x^2} \end{align*}, which is correct, as you need to simplify something in a square root. But since you are dealing with negative values of x, when you simplify this in the denominator, you will NOT be dividing by x, as you did, you should be dividing by \displaystyle \begin{align*} -x \end{align*}, since \displaystyle \begin{align*} \sqrt{x^2} = -x \end{align*} when x is negative.
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