Question

$\underset{x\to \infty }{\lim }\left[x\left\{\sqrt{x^2+1}-\sqrt{x^2-1}\right\}\right]$

Answer 1

$$\begin{gathered} \underset{x\to \infty }{\lim }\left[x\left\{\sqrt{x^2+1}-\sqrt{x^2-1}\right\}\right] \\ \mathrm{Rationalising}\mathrm{the}\mathrm{numerator}: \\ \underset{x\to \infty }{\lim }\left[x\left\{\frac{\left(\sqrt{x^2+1}-\sqrt{x^2-1}\right)\left(\sqrt{x^2+1}+\sqrt{x^2-1}\right)}{\left(\sqrt{x^2+1}+\sqrt{x^2-1}\right)}\right\}\right] \\ =\underset{x\to \infty }{\lim }\left[x\left\{\frac{\left(x^2+1\right)-\left(x^2-1\right)}{\left(\sqrt{x^2+1}+\sqrt{x^2-1}\right)}\right\}\right] \\ =\underset{x\to \infty }{\lim }\left[\frac{x\times 2}{\left(\sqrt{x^2+1}+\sqrt{x^2-1}\right)}\right] \\ \mathrm{Dividing}\mathrm{the}\mathrm{numerator}\mathrm{and}\mathrm{the}\mathrm{denominator}\mathrm{by}\mathrm{x}: \\ \underset{x\to \infty }{\lim }\left[\frac{2}{\sqrt{{\displaystyle \frac{x^2+1}{x^2}}}+\sqrt{{\displaystyle \frac{x^2-1}{x^2}}}}\right] \\ =\underset{x\to \infty }{\lim }\left[\frac{2}{\sqrt{{\displaystyle \frac{x^2+1}{x^2}}}+\sqrt{{\displaystyle \frac{x^2-1}{x^2}}}}\right] \\ =\underset{x\to \infty }{\lim }\left[\frac{2}{\sqrt{1+{\displaystyle \frac{1}{x^2}}}+\sqrt{1-{\displaystyle \frac{1}{x^2}}}}\right] \\ =\frac{2}{\sqrt{1}+\sqrt{1}} \\ =2 \end{gathered}$$