First, you have done the rationalisation a bit wrong. It should be

Next, think, in a non-rigorous way, about what you guess the answer ought to be. Notice that, when x is large, is very nearly , 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 ). Therefore 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

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 out of the square root. But remember that x is negative, so that the square root of will be –x, not x.