infinite limit question from Calc I

Question

Find the limit

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

This limit is part of a question involving squeeze theorum, the limit is $\frac12$ but i don't know how to prove it because of the polynomial in the radial, any help in apprechiated!

Answer 1

$$\eqalign{\lim_{x\to\infty}\sqrt{x^2+x+1}-x&=\lim_{x\to\infty}\frac{(\sqrt{x^2+x+1}-x)(\sqrt{x^2+x+1}+x)}{\sqrt{x^2+x+1}+x} \\&=\lim_{x\to\infty}\frac{x^2+x+1-x^2}{\sqrt{x^2+x+1}+x}\\&=\lim_{x\to\infty}\frac{x+1}{\sqrt{x^2+x+1}+x}\\&=\lim_{x\to\infty}\frac{1+1/x}{\sqrt{1+1/x+1/x^2}+1}\\&=\frac12.}$$

Answer 2

$$\displaystyle \lim_{x\to\infty} \sqrt{x^2+x+1}-x=\lim_{x\to\infty} \frac{(\sqrt{x^2+x+1}-x)(\sqrt{x^2+x+1}+x)}{\sqrt{x^2+x+1}+x}= \\ =\lim_{x\to\infty} \frac{x^2+x+1-x^2}{\sqrt{x^2+x+1}+x}=\lim_{x\to\infty} \frac{x+1}{\sqrt{x^2+x+1}+x}=\lim_{x\to\infty} \frac{1+\frac{1}{x}}{\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}+1}=\frac{1}{2}$$