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However x is negative, so, and
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The original expression should approach -1.
First of all, if you multiply top and bottom by the conjugate, you should be multiplying byOriginally Posted by kingsolomonsgrave
when I multiply byI get a denominator of zero. Does that mean I should divide thru by x to get
)
That does not seem right either., NOT
. Doing this gives
 &= \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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