Problem:

http://www.webassign.net/cgi-bin/sym...%202%20x%29%29

I rationalized it to:

-2/ (x-sqrt(x^2+2x))

What is the next step?? The book suggest dividing the numerator and denominator by x but I don't understand what to do afterwards.

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- Jul 7th 2011, 12:09 AMNeoSonataCompute Limit
Problem:

http://www.webassign.net/cgi-bin/sym...%202%20x%29%29

I rationalized it to:

-2/ (x-sqrt(x^2+2x))

What is the next step?? The book suggest dividing the numerator and denominator by x but I don't understand what to do afterwards. - Jul 7th 2011, 02:27 AMOpalgRe: Compute Limit
First, you have done the rationalisation a bit wrong. It should be $\displaystyle \frac{-2x}{x - \sqrt{x^2+2x}}.$

Next, think, in a non-rigorous way, about what you guess the answer ought to be. Notice that, when x is large, $\displaystyle x^2+2x$ is very nearly $\displaystyle (x+1)^2$, so its square root is approximately |x+1| = –x–1 (don't forget to take the absolute value there, because x+1 is negative as $\displaystyle x\to-\infty$). Therefore $\displaystyle x + \sqrt{x^2+2x}$ is approximately x + (–x–1) = –1. Thus you should expect that the limit is going to be –1. But how do you prove that?

Going back to the rationalised form, the book suggests dividing the numerator and denominator by x, which would give

$\displaystyle \frac{-2}{1 - \frac1x\sqrt{x^2+2x}} = \frac{-2}{1 - \frac1x\sqrt{x^2 \bigl(1+\frac2x}\bigr)}.$

That should be enough in the way of a hint, except that for the next step, you have to be careful. You want to take the factor $\displaystyle x^2$ out of the square root. But remember that x is negative, so that the square root of $\displaystyle x^2$ will be –x, not x. - Jul 7th 2011, 03:51 AMmr fantasticRe: Compute Limit