Question

Show that $\underset{x\to \infty }{\lim }\left(\sqrt{x^2+x+1}-x\right)\ne \underset{x\to \infty }{\lim }\left(\sqrt{x^2+1}-x\right)$

Answer 1

$$\begin{gathered} \underset{x\to \infty }{\lim }\left(\sqrt{x^2+x+1}-x\right)\ne \underset{x\to \infty }{\lim }\left(\sqrt{x^2+1}-x\right) \\ \mathrm{LHS}: \\ \underset{x\to \infty }{\lim }\left(\left(\sqrt{x^2+x+1}-x\right)\right) \\ \mathrm{Rationalising}\mathrm{the}\mathrm{numerator}: \\ \underset{x\to \infty }{\lim }\left[\frac{\left(\sqrt{x^2+x+1}-x\right)\left(\sqrt{x^2+x+1}+x\right)}{\left(\sqrt{x^2+x+1}+x\right)}\right] \\ =\underset{x\to \infty }{\lim }\left[\frac{\left(x^2+x+1\right)-x^2}{\left(\sqrt{x^2+x+1}+x\right)}\right] \\ =\underset{x\to \infty }{\lim }\left[\frac{x+1}{\left(\sqrt{x^2+x+1}+x\right)}\right] \\ \mathrm{Dividing}\mathrm{the}\mathrm{numerator}\mathrm{and}\mathrm{the}\mathrm{denominator}\mathrm{by}x: \\ \underset{x\to \infty }{\lim }\left[\frac{1+{\displaystyle \frac{1}{x}}}{{\displaystyle \frac{\sqrt{x^2+x+1}}{x}}+1}\right] \\ =\underset{x\to \infty }{\lim }\left[\frac{1+{\displaystyle \frac{1}{x}}}{\sqrt{{\displaystyle \frac{x^2+x+1}{x^2}}}+1}\right] \\ =\underset{x\to \infty }{\lim }\left[\frac{1+{\displaystyle \frac{1}{x}}}{\sqrt{1+{\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{x^2}}}+1}\right] \\ \mathrm{When}x\to \infty ,\mathrm{then}\frac{1}{x}\to 0. \\ \frac{1}{\sqrt{1}+1} \\ =\frac{1}{2} \\ \mathrm{RHS}: \\ \underset{x\to \infty }{\lim }\left(\sqrt{x^2+1}-x\right)\left[\mathrm{from}\infty -\infty \right] \end{gathered}$$

Rationalising the numerator:

$$\begin{gathered} \underset{x\to \infty }{\lim }\left[\frac{\left(\sqrt{x^2+1}-x\right)\left(\sqrt{x^2+1}+x\right)}{\left(\sqrt{x^2+1}+x\right)}\right] \\ =\underset{x\to \infty }{\lim }\left[\frac{x^2+1-x^2}{\left(\sqrt{x^2+1}+x\right)}\right] \\ =\frac{1}{\infty } \\ =0 \\ \therefore \underset{x\to \infty }{\lim }\left[\sqrt{x^2+x+1}-x\right]\ne \underset{x\to \infty }{\lim }\left(\sqrt{x^2+1}-x\right) \end{gathered}$$