<s>[INST] 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.
 [/INST] $$\lim_{x\to-\infty}\sqrt{ax^2+bx+c}=\lim_{x\to-\infty}\sqrt{a(x^2+\frac{b}{a}x+\frac{c}{a})}=$$
$$=\lim_{x\to-\infty}\sqrt{a}\times \sqrt{(x+\frac{b}{2a})^2+\frac{4ac-b^2}{4a}}=$$
$$=\lim_{x\to-\infty}\sqrt{a}\times \mid {x+\frac{b}{2a}}\mid\times\sqrt{1+\frac{\frac{4ac-b^2}{4a}}{(x+\frac{b}{2a})^2}}=$$
$$=\lim_{x\to-\infty}\sqrt{a}\times \mid {x+\frac{b}{2a}}\mid$$
$$=\lim_{x\to-\infty}\sqrt{a}\times (-x-\frac{b}{2a})$$
$$\lim_{x\to-\infty}(x-\sqrt{x^2+7x})=\lim_{x\to-\infty}(x-\sqrt{1}\times \mid {x+\frac{7}{2}}\mid)=$$
$$=\lim_{x\to-\infty}x-(-x-\frac{7}{2})=$$
$$=\lim_{x\to-\infty}(2x+\frac{7}{2})=-\infty$$
</s>