L'Hôpital won't help you this time!

We can substitute the (-1) in the denominator for (0), as it will not affect the final result (infinity minus one here equal infinity). By doing that, we have:

$lim_{x \to \infty}\frac{x}{\sqrt{x^2}}=lim_{x \to \infty} 1$
Because the limit of a constant is that constant, the limit of 1 as x approaches infinity is equal to $\boxed{1}$ $\square$