Limit of square root

Question

Can you help me to calculate this limit as '$a$' varies in $\mathbb{R}$: $$\lim _{x\rightarrow +\infty} \sqrt{2x^2 + x + 1} - ax$$

Answer 1

By multiplying numerator and denominator by $\sqrt{2x^2+x+1}+ax$ we obtain \begin{align*} \lim_{x\rightarrow+\infty}\frac{\sqrt{2x^2+x+1}-ax}{1}&=\lim_{x\rightarrow+\infty}\frac{2x^2+x+1-a^2x^2}{\sqrt{2x^2+x+1}+ax}\\ &=\lim_{x\rightarrow+\infty}\frac{x^2(2-a^2)+x+1}{x\left(\sqrt{2+\frac{1}{x}+\frac{1}{x^2}}+a\right)} \end{align*}

• If $a=\sqrt{2}$, then $$=\lim_{x\rightarrow+\infty}\frac{x\left(1+\frac{1}{x}\right)}{x\left(\sqrt{2+\frac{1}{x}+\frac{1}{x^2}}+\sqrt{2}\right)}=\lim_{x\rightarrow+\infty}\frac{1+\frac{1}{x}}{\sqrt{2+\frac{1}{x}+\frac{1}{x^2}}+\sqrt{2}}=\frac{1}{\sqrt{2}+\sqrt{2}}=\frac{1}{2\sqrt{2}}$$
• If $a=-\sqrt{2}$, then $$=\lim_{x\rightarrow+\infty}\frac{x\left(1+\frac{1}{x}\right)}{x\left(\sqrt{2+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{2}\right)}=\lim_{x\rightarrow+\infty}\frac{1+\frac{1}{x}}{\sqrt{2+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{2}}=\left[\frac{1}{0^+}\right]=+\infty$$
• If $a\neq \pm\sqrt{2}$, then $(2-a^2)\neq0$, so $$\lim_{x\rightarrow+\infty}\frac{x^2(2-a^2)+x+1}{x\left(\sqrt{2+\frac{1}{x}+\frac{1}{x^2}}+a\right)}=\lim_{x\rightarrow+\infty}\frac{x(2-a^2)+1+\frac{1}{x}}{\sqrt{2+\frac{1}{x}+\frac{1}{x^2}}+a}=\left[\frac{+\infty(2-a^2)+1}{\sqrt{2}+a}\right]=\pm\infty$$ You can notice, that
• if $a<-\sqrt{2}$, then $(2-a^2)<0$ and $\sqrt{2}+a<0$, so the limit is equal to $+\infty$
• if $a\in(-\sqrt{2},\sqrt{2})$, then $(2-a^2)>0$ and $\sqrt{2}+a>0$, so the limit is equal to $+\infty$
• if $a>\sqrt{2}$, then $(2-a^2)<0$ and $\sqrt{2}+a>0$, so the limit is equal to $-\infty$

Answer 2

For $a=\sqrt{2}$, the limit calculation is more complicated. $\sqrt{2x^2+x+1}=\sqrt{2}x\sqrt{1+\frac{x+1}{2x^2}}\approx\sqrt{2}x(1+\frac{1}{4x})$ Subtract $ax$ and get $\frac{\sqrt{2}}{4}$ as the limit.

For $a\ne \sqrt{2}$, this calculation leads easily to $\infty$ as the limit for $a\lt\sqrt{2}$ and $-\infty$ for $a\gt \sqrt{2}$.