Limit of $\sqrt{x^2+3x}+x$ when $x\to-\infty$ Limit of $ \lim_{x\to -\infty}(\sqrt{x^2+3x}+x)$, I know that the final answer is $-3/2$, my question is about Wolfram Alpha step by step solution:
$$x+\sqrt{x^2+3x}=\frac{(x+\sqrt{x^2+3x})(x-\sqrt{x^2+3x})}{x-\sqrt{x^2+3x}}$$
$$=-\frac{3x}{x-\sqrt{x^2+3x}}$$
$$\lim_{x\to-\infty}-\frac{3x}{x-\sqrt{x^2+3x}}$$
$$\lim_{x\to-\infty}-\frac{3x}{x-\sqrt{x^2+3x}}=-3$$
$$\lim_{x\to-\infty}\frac{x}{x-\sqrt{x^2+3x}}=-3\lim_{x\to-\infty}\frac{x}{x-\sqrt{x^2+3x}}$$
$$\frac{x}{x-\sqrt{x^2+3x}}=\frac{1}{1-\frac{\sqrt{x^2+3x}}{x}}$$
$$-3\lim_{x\to-\infty}\frac{x}{x\left(1-\frac{\sqrt{x^2+3x}}{x}\right)}$$
To prepare the product $\frac{1}{1-\frac{\sqrt{x^2+3x}}{x}}$ for solution by l'Hopital's rule, write it as $\frac{x}{x\left(1-\frac{\sqrt{x^2+3x}}{x}\right)}$
$$-3\lim_{x\to-\infty}\frac{x}{x\left(1-\frac{\sqrt{x^2+3x}}{x}\right)}$$

Is it correct to use L'Hopital here like Wolfram did?


You are right to render
$\sqrt{x^2+3x}+x=\frac{3x}{\sqrt{x^2+3x}-x}$
The next step is to complete the square on $x^2+3x$ getting
$x^2+3x=(x+(3/2))^2-(9/4)$
Use this to show that for negative $x$ with $x<-3$ (why?):
$-x-(3/2)<\sqrt{x^2+3x}<-x$
and then
$-\frac{3x}{2x+(3/2)}<\frac{3x}{\sqrt{x^2+3x}-x}<-(3/2)$
Now get the limit from the squeeze theorem.
