Evaluating $\lim\limits_{x \to -\infty} \sqrt{x^2 + 3x} - \sqrt{x^2 + x}$. Is Wolfram wrong or is it me? What am I doing wrong?
My attempt
$$\begin{align}
\lim_{x \to -\infty} \sqrt{x^2 + 3x} - \sqrt{x^2 + x} &= \lim_{x \to -\infty} \sqrt{x^2 + 3x} - \sqrt{x^2 + x} \cdot \frac{\sqrt{x^2 + 3x} + \sqrt{x^2 + x}}{\sqrt{x^2 + 3x} + \sqrt{x^2 + x}} =\\
&= \lim_{x \to -\infty} \frac{2x}{\sqrt{x^2 + 3x} + \sqrt{x^2 + x}} =\\
&= \lim_{x \to -\infty} \frac{\frac1x \cdot 2x}{\sqrt{\frac{x^2}{x^2} + \frac{3x}{x^2}} + \sqrt{\frac{x^2}{x^2} + \frac{x}{x^2}}} = \\
&= \lim_{x \to -\infty} \frac{2}{\sqrt{1 + \frac3x} + \sqrt{1 + \frac1x}} = 1
\end{align}$$
Wolfram result


Well very obviously Wolfram's answer is incoherent, since the constant term in the expansion at infinity and the limit should be equal. Your reasoning, however, fails on the last line, since $x\sqrt{a}$ does not equal $\sqrt{x^{2}a}$ as $x$ goes to minus infinity, but $-\sqrt{x^{2}a}$ instead.
