For what values of $a$ and $b$ does $\lim_{x\rightarrow \infty}(\sqrt{x^2+x+1}-ax-b)=1$?

Question

I have a question about limits tending to infinity. I need to find the constants $a$ and $b$ for which this limit takes the value 1. Please, help! Thank you!

$$\lim_{x\rightarrow \infty}(\sqrt{x^2+x+1}-ax-b)=1.$$

I've tried various things, including trial and error (it gave me the values $a=1$ and $b=\frac{1}{2}$), but it seems that I can't find a way to do this. Thank you!!

Answer 1

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

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

$\lim_{x\to \infty} \sqrt{x^2+x+1}-ax+b=1 \implies 1-b=\frac{1}{2} \implies b=\frac{1}{2}$

Answer 2

Here's an approach. Your limit is equivalent to $$\lim_{x\to\infty}(x(\sqrt{1+\frac1x+\frac1{x^2}}-a)-b)=1$$

Now, we can use a Taylor expansion, with the approximation $\frac1x\sim0$ of $(\sqrt{1+\frac1x+\frac1{x^2}}-a)$ to get the desired result.