Question

$\underset{x\to \infty }{\lim }\left(\sqrt{x^2+1}-x\right)$ is equal to _____________________________.

Answer 1

$$\begin{gathered} \begin{array}{c}\lim \\ x\to \infty \end{array}\left(\sqrt{x^2+1}-x\right) \\ \mathrm{i}.\mathrm{e}.\begin{array}{c}\lim \\ x\to \infty \end{array}\sqrt{x^2+1}-x\times \left(\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x}\right) \\ \mathrm{i}.\mathrm{e}.\begin{array}{c}\lim \\ x\to \infty \end{array}\frac{{\left(\sqrt{x^2+1}\right)}^2-x^2}{\sqrt{x^2+1}+x} \\ \mathrm{i}.\mathrm{e}.\begin{array}{c}\lim \\ x\to \infty \end{array}\frac{x^2+1-x^2}{\sqrt{x^2+1}+x}=\begin{array}{c}\lim \\ x\to \infty \end{array}\frac{x^2+1-x^2}{\sqrt{x^2+1}+x} \\ \mathrm{i}.\mathrm{e}.\begin{array}{c}\lim \\ x\to \infty \end{array}\frac{1}{\sqrt{x^2+1}+x}=\frac{1}{\infty }=0 \\ \mathrm{i}.\mathrm{e}.\begin{array}{c}\lim \\ x\to \infty \end{array}\left(\sqrt{x^2+1}-x\right)=0 \end{gathered}$$