Getting the wrong value from $\lim_\limits{x\to-\infty}x-\sqrt{x^2+7x}$ 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.
 A: 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.
A: The easiest way to see intuitively that your solution is wrong and that the answer is $- \infty$ is:
$$\lim_{x \to - \infty} \left (x-\sqrt{x^2+7x} \right) \leq \lim_{x \to -\infty} x = - \infty.$$
A: Thank you for all the answers. I will just leave a version of my own, with the help provided by the answers. Perhaps another straightforward way to do it is by changing the sign of $x$ right from the start
$$
\begin{align}
\lim_{x\to-\infty}x-\sqrt{x^2+7x}&=\lim_{x\to\infty}-x-\sqrt{x^2-7x}\\
&=-\lim_{x\to\infty}x-\lim_{x\to\infty}\sqrt{x^2-7x}\\
&=-\infty.
\end{align}
$$
A: $$\lim_{x\to-\infty}\sqrt{ax^2+bx+c}=\lim_{x\to-\infty}\sqrt{a(x^2+\frac{b}{a}x+\frac{c}{a})}=$$
$$=\lim_{x\to-\infty}\sqrt{a}\times \sqrt{(x+\frac{b}{2a})^2+\frac{4ac-b^2}{4a}}=$$
$$=\lim_{x\to-\infty}\sqrt{a}\times \mid {x+\frac{b}{2a}}\mid\times\sqrt{1+\frac{\frac{4ac-b^2}{4a}}{(x+\frac{b}{2a})^2}}=$$
$$=\lim_{x\to-\infty}\sqrt{a}\times \mid {x+\frac{b}{2a}}\mid$$
$$=\lim_{x\to-\infty}\sqrt{a}\times (-x-\frac{b}{2a})$$
$$\lim_{x\to-\infty}(x-\sqrt{x^2+7x})=\lim_{x\to-\infty}(x-\sqrt{1}\times \mid {x+\frac{7}{2}}\mid)=$$
$$=\lim_{x\to-\infty}x-(-x-\frac{7}{2})=$$
$$=\lim_{x\to-\infty}(2x+\frac{7}{2})=-\infty$$
