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?

 A: Yes, Wolfram is free to do that because for all finite $x$,
$$\frac{f(x)}{g(x)}=\frac{xf(x)}{xg(x)}$$ by "unsimplification", so that the limits are the same.
You can even write a "modified L'Hospital rule" theorem saying
$$\lim\frac fg=\lim\frac{f'}{g'}=\lim\frac{f+xf'}{g+xg'},$$ if that has any use.
A: 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.
A: $$
x+\sqrt{x^2+3x}=\frac{x^2-({x^2+3x})}{x-\sqrt{x^2+3x}}=\frac{-3x}{x-\sqrt{x^2+3x}}=\frac{-3}{1+\sqrt{1+\frac3x}}\stackrel{\tiny x\to-\infty}{\longrightarrow} -\frac 32
$$
A: $$\lim_{x\rightarrow-\infty}\left(\sqrt{x^2+3x}+x\right)$$
Multiply by the conjugate
\begin{aligned}
\sqrt{x^2+3x}+x &= \left(\sqrt{x^2+3x}+x\right) \cdot \left(\frac{\sqrt{x^2+3x}-x}{\sqrt{x^2+3x}-x}\right)\\
&= \frac{\left(\sqrt{x^2+3x}\right)^2-x^2}{\sqrt{x^2+3x}-x}\\
&= \frac{x^2+3x-x^2}{\sqrt{x^2+3x}-x}\\
&=\frac{3x}{\sqrt{x^2+3x}-x}
\end{aligned}
Also 
\begin{aligned}
\lim_{x\rightarrow-\infty}\left(\frac{3x}{\sqrt{x^2+3x}-x}\right)&= 3\cdot\lim_{x\rightarrow-\infty}\left(\frac{x}{\sqrt{x^2+3x}-x}\right)\\
&=3\cdot\lim_{x\rightarrow-\infty}\left(\frac{x}{\sqrt{x^2\left(1+\frac{3}{x}\right)}-x}\right)\\
&=3\cdot\lim_{x\rightarrow-\infty}\left(\frac{x}{\sqrt{x^2}\sqrt{1+\frac{3}{x}}-x}\right)
\end{aligned}
Let $x \to -\infty \implies \sqrt{x^2} = -x$
\begin{aligned}
3\cdot\lim_{x\rightarrow-\infty}\left(\frac{x}{-x\sqrt{1+\frac{3}{x}}-x}\right)
\end{aligned}
Divide by $x$
\begin{aligned}
3\cdot\lim_{x\rightarrow-\infty}\left(\frac{\frac{x}{x}}{-\frac{x\sqrt{1+\frac{3}{x}}}{x}-\frac{x}{x}}\right) = 3\cdot\lim_{x\rightarrow-\infty}\left( \frac{1}{-\sqrt{1+\frac{3}{x}}-1}\right)
\end{aligned}
\begin{aligned}
3\cdot \frac{\lim _{x\to -\infty }\left(1\right)}{\lim _{x\to -\infty }\left(-\sqrt{1+\frac{3}{x}}-1\right)} =3\cdot \frac{1}{-2} = -\frac{3}{2}
\end{aligned}
So

$$\lim_{x\rightarrow-\infty}\left(\sqrt{x^2+3x}+x\right) = -\frac{3}{2}$$



Note:
\begin{aligned}
\lim _{x\to -\infty }\left(-\sqrt{1+\frac{3}{x}}-1\right) = \lim _{x\to -\infty }\left(-\sqrt{1+\frac{3}{x}}\right)-\lim _{x\to -\infty }\left(1\right)
= 1 - (-1) =2
\end{aligned}
