We know that
lim_{x \to -\infty} [ sqrt(x^2 + x) - x ] = lim_{x \to -\infty} sqrt(x^2 + x) + lim_{x \to -\infty} [ - x ] ,
and, trivially,
lim_{x \to -\infty} [- x] = \infty.
Therefore, the interesting part is the limit of the square root. If we look at the term under this square root, then, if x goes to -\infty, we have a positive part, x^2, and a negative part, x. However, as x^2 grows much faster than x decays, we know that the limit has to be \infty. We can prove this by showing that for an arbitrary \epsilon > 0 it holds that sqrt(x^2 + x) > \epsilon if x is chosen small enough (which is guaranteed to be the case at some point, as x goes to -\infty).
Starting the proof, we set
sqrt(x^2 + x) > \epsilon,
and thus get
x^2 + x > \epsilon^2.
By completing the square, we find that this is the same as
(x + 1/2)^2 - 1/4 > \epsilon^2,
and hence get
(x + 1/2)^2 > \epsilon^2 + 1/4
or
| x + 1/2 | > \sqrt(\epsilon^2 + 1/4).
Finally, noting that | x + 1/2 | >= | x | - 1 / 2, we arrive at
| x | > \sqrt(\epsilon^2 + 1/4) + 1/2 = {\sqrt{4 \epsilon^2 + 1} + 1} / 2.
This shows that
lim_{x \to -\infty} sqrt(x^2 + x) = \infty,
and it follows that
lim_{x \to -\infty} [ sqrt(x^2 + x) - x ] = \infty.