# 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

• Possibly you make an error when calculating a square root. Most common mistake is taking $\sqrt{x^2} = x$ even if $x$ is negative. – Hanul Jeon Apr 24 '16 at 13:36
• Argh! Never write anything like that first line! – Henning Makholm Apr 24 '16 at 13:37
• @Henning Why not? It would be a useful way of determining what kind of indeterminate form you're dealing with, as well as making sure that it can't be decided by pure substitution (as long as you make sure to stop at $\infty-\infty$). People write things like $\frac\infty\infty$ and $1^\infty$ all the time. – Arthur Apr 24 '16 at 13:41
• @Arthur: I'm objecting to the equals signs, not to shorthand classification of indeterminate limits in general. – Henning Makholm Apr 24 '16 at 13:44
• @HenningMakholm Fair enough. I agree that $=$ is the wrong symbol. – Arthur Apr 24 '16 at 13:46

On your way to the last line, you're tacitly assuming that, for example, $\frac1x\sqrt{x^2+x} = \sqrt{\frac{x^2}{x^2}+\frac{x}{x^2}}$. But that is only true when $x$ is positive!

When $x$ is negative, $\frac1x\sqrt{\cdots\vphantom{x}}$ will be negative, and thus it can never be written as a sum of square roots.

• Thank you very much. sorry, I dont know how to write here math things so, always when I'm gonna put some "x" inside de sqrt, if x is negative, I need to assume that (1/x)*sqrt(x^2) = - sqrt (1) = -1 no? thank you so much – user334236 Apr 24 '16 at 13:43

Your error is when you divided the numerator by $x$ but you divided the denominator by $\sqrt{x^2}$. Note that when $x$ is negative, $\sqrt{x^2}$ is positive.

• So, since $x<0$, when you divided the denominator by $\sqrt{x^2}$, you should have divided the numerator by $-x$. – Joel Reyes Noche Apr 24 '16 at 13:46

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.