Show that lim x--> infinity square root (x^2+x) -x = 1/2

note the x^2 + x is under the root only

Here's what I did so far:

[square root (x^2+x) -x] [square root (x^2+x) +x] / [square root (x^2+x) +x]

If we expand it out it becomes:

x^2+x + x[square root (x^2+x)] - x[square root (x^2+x)] - x^2 / [square root (x^2+x) +x]

1/ [square root (x^2+x) +x]

annnd that's where i lose it.. (if i ever had it from the start)