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Math Help - for this limit problem do I multiply by the conjugate?

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    for this limit problem do I multiply by the conjugate?

    for this limit problem do I multiply by the conjugate?-picture-22.png

    when I multiply by x+sqrt(x^2+2x) I get a denominator of zero. Does that mean I should divide thru by x to get 1+sqrt(1+2)

    That does not seem right either.
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    Re: for this limit problem do I multiply by the conjugate?

    x + \sqrt{x^2 + 2x} = x + \sqrt{(x+1)^2 - 1} < x + \sqrt{(x+1)^2}

    However x is negative, so \sqrt{(x+1)^2} = -(x+1), and x - (x+1) = -1.

    The original expression should approach -1.
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    Re: for this limit problem do I multiply by the conjugate?

    Quote Originally Posted by kingsolomonsgrave View Post
    Click image for larger version. 

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    when I multiply by x+sqrt(x^2+2x) I get a denominator of zero. Does that mean I should divide thru by x to get 1+sqrt(1+2)

    That does not seem right either.
    First of all, if you multiply top and bottom by the conjugate, you should be multiplying by \displaystyle \begin{align*} x - \sqrt{x^2 + 2x} \end{align*}, NOT \displaystyle \begin{align*} x + \sqrt{x^2 + 2x} \end{align*}. Doing this gives

    \displaystyle \begin{align*} \lim_{ x \to -\infty}\left( x + \sqrt{x^2 + 2x} \right) &= \lim_{x \to -\infty}\frac{\left( x + \sqrt{x^2 + 2x} \right) \left( x - \sqrt{x^2 + 2x} \right) }{x - \sqrt{x^2 + 2x} } \\ &= \lim_{x \to -\infty} \frac{x^2 - \left( x^2 + 2x \right) }{x - \sqrt{x^2 + 2x} } \\ &= \lim_{x \to -\infty} \frac{-2x}{x - \sqrt{x^2 + 2x}} \\ &= \lim_{x \to -\infty} \frac{\frac{1}{|x|} \left( -2x \right) }{\frac{1}{|x|} \left( x - \sqrt{x^2 + 2x} \right)} \\ &= \lim_{x \to -\infty} \frac{\frac{1}{-x} \left( -2x \right) }{ \frac{1}{-x} \left(x \right) - \frac{1}{\sqrt{(-x)^2}} \sqrt{ x^2 + 2x } } \textrm{ since we know that } x < 0 \\ &= \lim_{ x\to -\infty} \frac{2}{-1 - \sqrt{\frac{x^2 + 2x}{(-x)^2}}} \\ &= \lim_{x \to -\infty} \frac{2}{-1 - \sqrt{1 + \frac{2}{x}}} \\ &= \frac{2}{-1 - \sqrt{1 + 0}} \\ &= \frac{2}{-1 - 1} \\ &= -1  \end{align*}
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