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.

Answer 2

$$\lim_{x\to-\infty}\sqrt{ax^2+bx+c}=\lim_{x\to-\infty}\sqrt{a(x^2+\frac{b}{a}x+\frac{c}{a})}=$$ $$=\lim_{x\to-\infty}\sqrt{a}\times \sqrt{(x+\frac{b}{2a})^2+\frac{4ac-b^2}{4a}}=$$ $$=\lim_{x\to-\infty}\sqrt{a}\times \mid {x+\frac{b}{2a}}\mid\times\sqrt{1+\frac{\frac{4ac-b^2}{4a}}{(x+\frac{b}{2a})^2}}=$$ $$=\lim_{x\to-\infty}\sqrt{a}\times \mid {x+\frac{b}{2a}}\mid$$ $$=\lim_{x\to-\infty}\sqrt{a}\times (-x-\frac{b}{2a})$$

$$\lim_{x\to-\infty}(x-\sqrt{x^2+7x})=\lim_{x\to-\infty}(x-\sqrt{1}\times \mid {x+\frac{7}{2}}\mid)=$$ $$=\lim_{x\to-\infty}x-(-x-\frac{7}{2})=$$ $$=\lim_{x\to-\infty}(2x+\frac{7}{2})=-\infty$$

Answer 3

The easiest way to see intuitively that your solution is wrong and that the answer is $- \infty$ is:

$$\lim_{x \to - \infty} \left (x-\sqrt{x^2+7x} \right) \leq \lim_{x \to -\infty} x = - \infty.$$

Answer 4

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 $$ \begin{align} \lim_{x\to-\infty}x-\sqrt{x^2+7x}&=\lim_{x\to\infty}-x-\sqrt{x^2-7x}\\ &=-\lim_{x\to\infty}x-\lim_{x\to\infty}\sqrt{x^2-7x}\\ &=-\infty. \end{align} $$