Evaluate: $\lim\limits_{x \to \infty} \; \left[x-\sqrt{x^2+x}\right]$
Let $x = \dfrac{1}{y}$
Then, as $x \to \infty, \; y \to 0$
$\begin{aligned}
\therefore \lim\limits_{x \to \infty} \; \left[x-\sqrt{x^2+x}\right] & = \lim\limits_{y \to 0} \; \left[\dfrac{1}{y}-\sqrt{\dfrac{1}{y^2}+\dfrac{1}{y}}\right] \\
& = \lim\limits_{y \to 0} \; \left[\dfrac{1}{y}-\dfrac{\sqrt{1+y}}{y}\right] \\
& = \lim\limits_{y \to 0} \; \left[\dfrac{1-\sqrt{1+y}}{y}\right] \\
& = \lim\limits_{y \to 0} \; \left[\dfrac{\left(1-\sqrt{1+y}\right) \left(1+\sqrt{1+y}\right)}{y \left(1+\sqrt{1+y}\right)}\right] \\
& = \lim\limits_{y \to 0} \; \dfrac{1-1-y}{y\left(1+\sqrt{1+y}\right)} \\
& = \lim\limits_{y \to 0} \; \dfrac{-1}{1+\sqrt{1+y}} \\
& = \dfrac{-1}{1+\sqrt{1+0}} = \dfrac{-1}{2}
\end{aligned}$