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?


$$
x+\sqrt{x^2+3x}=\frac{x^2-({x^2+3x})}{x-\sqrt{x^2+3x}}=\frac{-3x}{x-\sqrt{x^2+3x}}=\frac{-3}{1+\sqrt{1+\frac3x}}\stackrel{\tiny x\to-\infty}{\longrightarrow} -\frac 32
$$
