# Puzzled by $\lim\limits_{x \to - \infty} \sqrt{x^2+x}-x$

I am preparing for the next Semester and therefore review a few of my Analysis I limits, I have found this example in C.T. Michaels Analysis I:

Compute $\displaystyle \lim_{x \to - \infty} \sqrt{x^2+x}-x$

Aprior to this exercise I computed the same limit but as $x$ approaches $\infty$ rather than $- \infty$. So I thought that this should be a piece of cake, but apparently the $- \infty$ makes all the difference for me.

My approach: This is the general approach I take when it comes to roots, especially square roots. Consider: $$\sqrt{x^2+x}-x= \left(\sqrt{x^2+x}-x\right)\cdot \frac{\sqrt{x^2+x}+x}{\sqrt{x^2+x}+x}= \frac{x^2+x-x^2}{\sqrt{x^2+x}+x}=\frac{x}{\sqrt{x^2+x}+x}$$ Factoring out an $x$ will get me to: $$\frac{x}{x\left(\sqrt{1+\frac{1}{x}}+1\right)}=\frac{1}{\sqrt{1+\frac{1}{x}}+1}$$ So as I take the limit of the above expressing as $x$ approaches $\infty$ I obtain the correct answer $1/2$.

However, when I study the limit as $x$ approaches $- \infty$ I don't see how that would make a difference since $1 / - \infty=0$, but the correct answer in that case would be $\infty$ http://www.wolframalpha.com/input/?i=lim+x+to+-infty+sqrt%28x%5E2%2Bx%29-x

My question(s):

Where is/are my mistakes?

Is it not possible to use the same methods for $- \infty$ as for $\infty$ when studying limits?

• Well, it might have been easier to just replace $x \rightarrow -x$ and find $\lim_{x \rightarrow \infty} \sqrt{x^2-x}+x=\infty$. Jan 5, 2014 at 9:09
• Similar limit: math.stackexchange.com/questions/603628/… Jan 5, 2014 at 9:55

The difficulty here is in "factoring out" the $x$. What you really did there is this:

You claim that $x^2+x=x^2(1+\frac{1}{x})$, and that $\sqrt{x^2(1+\frac{1}{x})}=\sqrt{x^2}\sqrt{1+\frac{1}{x}}$. This is perfectly fine so far, as long as the latter square root is defined. The problem is in the last step: you've claimed, from here, that $\sqrt{x^2}=x$. And that's not true!

Remember: in general, $\sqrt{x^2}=\lvert x\rvert$, not $x$. And if $x<0$, then $\lvert x\rvert=-x$, not $x$.

So, it should be $$\sqrt{x^2+x}=\sqrt{x^2}\sqrt{1+\frac{1}{x}}=\lvert x\rvert\sqrt{1+\frac{1}{x}}=-x\sqrt{1+\frac{1}{x}}.$$

• Here as well, great answer and very detail rich. I appreciate your effort a lot, I clearly underestimated this problem but you've brought clarity into it. Thanks. Jan 4, 2014 at 22:21

If $x$ is negative, then the square root must be manipulated differently; that is, we have

$$\sqrt{x^2 + x} = \sqrt{x^2} \sqrt{1 + \frac 1 x} = |x| \sqrt{1 + \frac 1 x}$$

But since $x < 0$, $|x| = -x$. This leads us to taking the limit of

$$\frac{x}{-|x|\sqrt{1 + \frac 1 x} + x} = \frac{1}{-\sqrt{1 + \frac 1 x} + 1}$$

Letting $x \to -\infty$, the denominator tends to zero.

• Much appreciated, two very clear and strong answers by you and @Nicholas R. Peterson. To keep things as fair as possible, I will up vote this one and accept Nicholas' answer. Again, thanks a lot. Jan 4, 2014 at 22:17

In addition to all the other answers, just note that if we set

$y = -x$ then we have

$$\lim_{y \to +\infty} \sqrt{y^2 - y} + y$$ which is clearly $+\infty$

• Thanks for showing me your approach @Ant, you're right, using this and asymptotic behavior it becomes very easy to see that the limit is $+ \infty$, I really appreciate this. Jan 7, 2014 at 14:45

Setting $\displaystyle-\frac1x=h\iff x=-\frac1h$

$$\lim_{x\to-\infty}\left(\sqrt{x^2+x}-x\right)=\lim_{h\to0^+}\left(\sqrt{\frac1{h^2}-\frac1h}+\frac1h\right)$$ $$=\lim_{h\to0^+}\frac{\sqrt{1-h}+1}h\text{ as }h>0$$

$$=\lim_{h\to0^+}\frac{(1-h)-1}{h(\sqrt{1-h}-1)}$$

$$=-\lim_{h\to0^+}\frac1{\sqrt{1-h}-1}\text{ as } h\ne0\text{ as }h\to0^+$$

$$=-\frac1{1-1}$$

• I really like this approach as well, thanks a lot for your time and effort to share this with me, I appreciate it. Thank you. Jan 5, 2014 at 22:14

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\mbox{Use always}\ \root{x^{2}} = \verts{x}\quad \mbox{and}\quad \verts{x} = x\,\sgn\pars{x}\ !}$ $${x \over \root{x^{2} + x} + x} = {\verts{x} \over \sgn\pars{x}\root{x^{2} + x} + \verts{x}} = {1 \over \sgn\pars{x}\root{1 + 1/x} + 1}\,,\qquad x \not= 0$$

\begin{align} &\lim_{x \to -\infty}{1 \over \sgn\pars{x}\root{1 + 1/x} + 1} = \lim_{x \to -\infty}{1 \over\ -\root{1 + 1/x} + 1} = +\infty \\[3mm]& \lim_{x \to +\infty}{1 \over \sgn\pars{x}\root{1 + 1/x} + 1} = \lim_{x \to +\infty}{1 \over\ +\root{1 + 1/x} + 1} = \half \end{align}

Esentially, your error was the division $\ds{\root{x^{2} + x} \over x}$. It's, indeed: $${\root{x^{2} + x} \over x} = -\,{\root{x^{2} + x} \over \verts{x}}\ \mbox{when}\ x < 0\quad\mbox{and}\quad {\root{x^{2} + x} \over x} = +\,{\root{x^{2} + x} \over \verts{x}}\ \mbox{when}\ x > 0$$