Calculate $\lim_{x \to \infty} \frac{1}{\sqrt{x^2+x+1}}$

Question

So it is obvious that this limit goes to $0$. I can reason that it does, say some hand-wavy things like "The bottom goes to infinity while top stays at $1$ and so the limit is $0$", but there is no rigor in that. If I wanted a rigid proof would I just apply the squeeze theorem? squeeze it between $0$ and $\frac{1}{x^2}$? Or is there an easier more straightforward way.

Thanks.

Answer 1

HINT:

$$\lim\limits_{x\to\pm\infty}\frac 1{x^n}=0$$

Can you finish the rest? I have provided the hidden answer below. Look if you are truly stuck!

$$\begin{align*}L=\lim\limits_{x\to\infty}\frac 1{\sqrt{1+x+x^2}} & =\lim\limits_{x\to\infty}\frac {\tfrac 1x}{\sqrt{\frac 1{x^2}+\frac 1x+1}}\\ & =\frac {0}{\sqrt{0+0+1}}\\ & =0\end{align*}$$

Answer 2

Note that $\frac1{\infty}$ is not an indeterminate form thus we can conlcude by algebraic theorems, as an alternative by squeeze theorem

$$0 \le \frac{1}{\sqrt{x^2+x+1}}\le \frac{1}{\sqrt{x^2}}=\frac{1}{x}\to0$$

and the last can be easily proved by $\epsilon$-$\delta$ definition.

Answer 3

Let $\varepsilon>0$, then for $x>\frac1{\varepsilon}$ we have $$0<\frac1{\sqrt{x^2+x+1}}<\frac1{\sqrt{x^2}}=\frac1x<\varepsilon$$