Solving this limit without l'Hôpital $\lim_{x \to -\infty} \sqrt{x^2 +x + 4} + x$

Question

I'm trying to solve this limit $\ \lim_{x \to -\infty} \sqrt{x^2 +x + 4} + x $. I tried solving it by calculating the limit of $\sqrt{x^2 +x + 4}$ and $ \ x $ in apart from eachother and simply adding them, however this would get me 0 and wolfram doesn't seem to agree with me on that one. I also tried to find this limit by multiplying by it's conjugate expression which ended up with me trying to divide by zero, so I doubt I found the solution there. Any help would be greatly appreciated

Answer 1

As is typical in exercises involving differences with square roots, an elementary method is the application of binomial theorems: $$ \sqrt{x^2+x+4}+x = \frac{(x^2+x+4)-x^2}{\sqrt{x^2+x+4}-x}=\frac{-|x|+4}{\sqrt{x^2-|x|+4}+|x|}, $$ the last equality is valid for $x<0$.

Answer 2

Set $y=-1/x$ to get $$x^2+x+4=\dfrac{1-y+4y^2}{y^2}$$

$$\sqrt{x^2+x+4}=+\dfrac{\sqrt{1-y+4y^2}}{|y|}$$

as $y>0, |y|=+y$

$$\lim_{x\to-\infty}\sqrt{x^2+x+4}-x=\lim_{y\to0^+}\dfrac{\sqrt{1-y+4y^2}-1}y$$

Now rationalize the numerator