Getting the wrong value from $\lim_\limits{x\to-\infty}x-\sqrt{x^2+7x}$

Question

Consider the following limit
$$
\lim_{x\to-\infty}x-\sqrt{x^2+7x}
$$
Going through some algebra leads to
$$\begin{align}
\lim_{x\to-\infty}x-\sqrt{x^2+7x}&=\lim_{x\to-\infty}\frac{(x-\sqrt{x^2+7x})(x+\sqrt{x^2+7x})}{(x+\sqrt{x^2+7x})}\\
&=\lim_{x\to-\infty}\frac{-7x}{x+\sqrt{x^2+7x}}\\
&=\lim_{x\to-\infty}\frac{-7}{1+\sqrt{1+7/x}}=-\frac72
\end{align}$$
Using WolframAlpha's step-by-step solution, however, gives this limit to be $-\infty$. Here's what it's doing
$$\begin{align}
\lim_{x\to-\infty}x-\sqrt{x^2+7x}&=\lim_{x\to-\infty}x-\lim_{x\to-\infty}\sqrt{x^2+7x}
\end{align}$$
where, by the power rule,
$$\begin{align}
\lim_{x\to-\infty}\sqrt{x^2+7x}&=\sqrt{\lim_{x\to-\infty}(x^2+7x)}\\
&=\sqrt{\lim_{x\to-\infty}x^2}\\
&=\sqrt{\left(\lim_{x\to-\infty}x \right)^2}=\infty
\end{align}$$
and so
$$
\lim_{x\to-\infty}x-\lim_{x\to-\infty}\sqrt{x^2+7x}=-\infty-\infty=-\infty
$$
What is wrong in this solution? I wondered if it was the power rule failing for undefined values of the square root, but not sure what to argue.

Answer 1

Your computation has an "absolute value" error.

One of your steps uses the equation $$\frac{\sqrt{x^2+7x}}{x} = \sqrt{\frac{x^2+7x}{x^2}} = \sqrt{1 + \frac{7}{x}} $$ However, this equation is only valid if $x > 0$, and you have applied it when $x < 0$ (as $x \to -\infty$).

In the case where $x < 0$, and therefore $x = - |x|$, what you could have done is this: $$\frac{\sqrt{x^2+7x}}{x} = -\frac{\sqrt{x^2+7x}}{|x|} = - \sqrt{\frac{x^2+7x}{x^2}} = - \sqrt{1 + \frac{7}{x}} $$ If you had done that, then your solution would have gone on to give a $0$ in the denominator, making the limit expression invalid.