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

1. ## for this limit problem do I multiply by the conjugate?

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.

2. ## 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.

3. ## Re: for this limit problem do I multiply by the conjugate?

Originally Posted by kingsolomonsgrave

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*}