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