Answer
The solution is
$$\lim_{x\to\infty}(x-\sqrt{x^2+4x})=-2.$$
Work Step by Step
To solve this limit follow the steps below
$$\lim_{x\to\infty}(x-\sqrt{x^2+4x})=\lim_{x\to\infty}(x-\sqrt{x^2+4x})\cdot\frac{x+\sqrt{x^2+4x}}{x+\sqrt{x^2+4x}}=\lim_{x\to\infty}\frac{x^2-\sqrt{x^2+4x}^2}{x+\sqrt{x^2+4x}}=\lim_{x\to\infty}\frac{x^2-x^2-4x}{x+\sqrt{x^2+4x}}=\lim_{x\to\infty}\frac{-4x}{x+\sqrt{x^2+4x}}=\lim_{x\to\infty}\frac{-4x}{x\left(1+\sqrt{1+\frac{4}{x}}\right)}=\lim_{x\to\infty}\frac{-4}{1+\sqrt{1+\frac{4}{x}}}=\left[\frac{-4}{1+\sqrt{1+\frac{4}{\infty}}}\right]=\left[\frac{-4}{1+\sqrt{1+0}}\right]=-2.$$
