Find the constant $p$ and $q$ in the $\lim_{x \to \infty}$

Question

Given $$\lim_{ x\to \infty} \sqrt{x^2-x+1} -px-q=0$$ find the values of $p$ and $q$. The answer is $p=1$ and $q= -1/2$.

I know the method is using L'Hospital’s rule and I rationalise the denominator. But I don’t know how to find the p and q. Thank you very much!

Answer 1

HINT

We have that

$$\sqrt{x^2-x+1}-px-q=\left[\sqrt{x^2-x+1}-(px+q)\right]\frac{\sqrt{x^2-x+1}+(px+q)}{\sqrt{x^2-x+1}+(px+q)}=$$

$$=\frac{x^2-x+1-(px+q)^2}{\sqrt{x^2-x+1}+(px+q)}=\frac{x^2-x+1-p^2x^2-2pqx-q^2}{\sqrt{x^2-x+1}+(px+q)}$$

Answer 2

By completing the square,

$$\sqrt{x^2-x+1}=\sqrt{\left(x-\frac12\right)^2+\frac34}=\left(x-\frac12\right)\sqrt{1+\frac3{4{\left(x-\dfrac12\right)^2}}}$$ hints you the values of $p$ and $q$ (as the second factor quickly converges to $1$). Then you can check by rationalizing the numerator.

Answer 3

The fastest method consists in a substitution $x=\dfrac1u$ $\;(u\to 0_+)$ to obtain $$f(x)=\frac1u\sqrt{1-u+u^2\strut},$$ and taking Taylor's expansion at order $2$ for the square root near $0$: $$\sqrt{1-u+u^2\strut}=1+\frac12(-u+u^2)-\frac18(-u+u^2)^2+o(u^2)=1-\frac12u+\frac38u^2+o(u^2),$$ so $$f(x)=\frac1u-\frac12+\frac38u+o(u)=x-\frac12+\frac3{8x}+o\Bigl(\frac1x\Bigr).$$

Thus we obtain $p=1$, $\;q=-\dfrac12$, and we have one more information: near $+\infty$, $$f(x)-\Bigl(x-\frac12\Bigr)>0$$ since $\frac3{8x}>0$.