$\lim_{x\to -\infty}\sqrt{x^2+5x+3}+x = \lim_{x\to -\infty}(-x)+x=\lim_{x\to -\infty}0 = 0$

Apparently, the 2nd step is illegal here. Probably because for $x=-\infty$ I'd get $(+\infty-\infty)$ which is not possible. I see why this wouldn't be possible, I'm not sure if it really is the cause which makes that equation illegal though.

But now, apparently, I could do this:

$\lim_{x\to -\infty}\sqrt{x^2+5x+3}+x=\lim_{x\to -\infty}\frac{[\sqrt{x^2+5x+3}+x][\sqrt{x^2+5x+3}-x]}{\sqrt{x^2+5x+3}-x}=\lim_{x\to -\infty}\frac{x^2+5x+3-x^2}{\sqrt{x^2+5x+3}-x}=\lim_{x\to -\infty}\frac{5x+3}{\sqrt{x^2+5x+3}-x}=\lim_{x\to -\infty}\frac{5+3/x}{\sqrt{1+5/x+3/x^2}-1}=-5/2$

which gives me the correct result.

But in the 3th step I used $x^2-x^2=0$, how is that legal? Also, in the 2nd step I implicitly used:


Which also seems to be fine, but why?

  • $\begingroup$ What's the issue with $x^2-x^2=0$? It's a perfectly legal statement. One shouldn't doubt oneself too much. On the other hand the issue with your first attempt is that you can't replace $\sqrt{x^2+5x+3}$ with $-x$ as they are not equal. One can only replace one expression with another expression when both are equal. $\endgroup$
    – Paramanand Singh
    Dec 4, 2018 at 3:57
  • $\begingroup$ Technically, your first two inequalities are not wrong because all terms equal 0. $\endgroup$
    – Dirk
    Dec 4, 2018 at 5:55
  • $\begingroup$ How did you even make that huge leap of faith to go from $\sqrt{x^2+5x+3}$ to $-x$? Even if you made $\sqrt{x^2+5x+3}$ become $x$, although wrong, this would still be understandable, but where did the $-$ pop up from? $\endgroup$
    – YiFan Tey
    Dec 5, 2018 at 2:55
  • $\begingroup$ @YiFan: you should note that $x$ is negative and square root is always non-negative so $-x$ is a reasonable choice. $\endgroup$
    – Paramanand Singh
    Dec 5, 2018 at 3:34

7 Answers 7


$\lim_{x\to -\infty}\sqrt{x^2+5x+3}+x = \lim_{x\to -\infty}(-x)+x$

Apparently, the 2nd step is illegal here.

In this step, you dropped $5x+3$ to go from $\sqrt{x^2}$ to $-x$, but what makes you think you can just leave out the terms $5x+3$...? Surely you agree that $x^2+5x+3 \ne x^2$.

But in the 3th step I used $x^2-x^2=0$, how is that legal?

How could that not be legal? The difference of two equal real numbers is $0$, of course!

The same goes for:

Also, in the 2nd step I implicitly used: $-x\sqrt{x^2+5x+3}+x\sqrt{x^2+5x+3}=0$

Addition after comment: the idea of dominating terms (in a polynomial) is fine, but you cannot choose to apply the limit "locally" and leave the other terms unchanged.

To clarify, this is fine: $$\lim_{x \to -\infty}\sqrt{x^2+5x+3}=\lim_{x \to -\infty}\sqrt{x^2}=+\infty$$ but that's not what we have here; we have: $$\lim_{x \to -\infty}\left(\sqrt{x^2+5x+3}\color{blue}{+x}\right)$$ and we cannot take the limit of the terms separately $$\lim_{x \to -\infty}\left(\sqrt{x^2+5x+3}\color{blue}{+x}\right)\color{red}{\ne}\lim_{x \to -\infty}\sqrt{x^2+5x+3}+\lim_{x \to -\infty} x$$ because both limits need to exist in order for this step to be valid.

  • $\begingroup$ The idea behind $\lim_{x\to -\infty}\sqrt{x^2+5x+3}=\lim_{x\to -\infty}\sqrt{x^2}=\lim_{x\to -\infty} |x| = \lim_{x\to -\infty} -x$ was, that $x^2$ dominantes $5x+3$. I also thought if that argument is wrong here, if $\lim_{x \to -\infty} x-x=0$ is legal and everything of the same "type"; then I'm okay again. Then I have to think about why the domination argument doesn't work as I thought it does here. $\endgroup$
    – xotix
    Dec 3, 2018 at 14:39
  • $\begingroup$ @xotix I added a part; the problem lies in taking the limit(s) separately/"locally". $\endgroup$
    – StackTD
    Dec 3, 2018 at 14:45

The first step you took simply wasn’t correct.

$$\lim_{x \to -\infty}\sqrt{x^2+5x+3}+x \color{red}{\neq \lim_{x \to \infty}-x+x}$$

You shouldn’t split a limit if the individual limits don’t exist or are undefined. Going for $\infty-\infty$ is certainly incorrect as it’s an indeterminate form, so you should always avoid it.

As for your question, you’re cancelling out $x^2$ and $-x^2$. That’s simple algebra and has nothing to do with whether $x$ is approaching an infinite value or not. It always applies for all additive inverses.


You may set $x = -\frac{1}{t}$ and consider the limit for $\stackrel{t\rightarrow 0+}{\longrightarrow}$: $$\begin{eqnarray*} \sqrt{x^2+5x+3}+x & \stackrel{x = -\frac{1}{t}}{=} & \frac{\sqrt{1-5t +3t^2} - 1}{t} \\ & \stackrel{t\rightarrow 0+}{\longrightarrow} & f'(0) = -\frac{5}{2}\mbox{ for } f(t) = \sqrt{1-5t +3t^2} \end{eqnarray*}$$


The first step is illegal, not the second. This is because $\sqrt{x^2 + 5x + 3} \not\equiv -x$.

Also addressing your comment:

The idea behind $\lim_{x \to -\infty} \sqrt{x^2 + 5x + 3} = \dots = \lim_{x \to -\infty} −x$ was, that $x^2$ dominantes $5x+3$. I also thought if that argument is wrong here, if $\lim_{x \to -\infty} x − x = 0$ is legal and everything of the same "type"; then I'm okay again. Then I have to think about why the domination argument doesn't work as I thought it does here.

You are right in thinking that $x^2$ dominates $5x+3$, so that $\color{blue}{\sqrt{x^2 + 5x + 3}} \sim \color{red}{-x}$ for large negative $x$. Indeed there is no problem when computing their ratio:

$$\lim_{x \to -\infty} \frac{\color{blue}{\sqrt{x^2 + 5x + 3}}}{\color{red}{-x}} = 1$$

However, because you are computing the difference between $\color{blue}{\sqrt{x^2 + 5x + 3}}$ and $\color{red}{-x}$, such an approximation is not fine enough. You need more terms of the series expansion:

$$ \begin{align*} \color{blue}{\sqrt{x^2 + 5x + 3}} &= -x \sqrt{1 + \frac{5}{x} + \frac{3}{x^2}} \\ &\sim -x \left(1 + \frac{1}{2} \cdot \frac{5}{x} + O\left(\frac{1}{x^2}\right) \right) \\ &= \color{red}{-x} - \frac{5}{2} + O\left(\frac{1}{x}\right) \end{align*} $$

Then may you conclude that the required limit is $-5/2$.

By analogy, an approximation such as $\color{blue}{x + 1} \sim \color{red}{x}$ is perfectly fine for computing the ratio limit $\lim_{x \to -\infty} (\color{blue}{x + 1}) / (\color{red}{x})$. But if you are computing the difference limit, then you may not discard the $1$ in $\color{blue}{x + 1}$.


To avoid confusion with sign, in these cases I suggest to take $y=-x\to \infty$ then

$$\lim_{x\to -\infty}\sqrt{x^2+5x+3}+x =\lim_{y\to \infty}\sqrt{y^2-5y+3}-y $$

from here we can clearly see that the expression is an indeterminate form $\infty-\infty$ then we can proceed by

$$\sqrt{y^2-5y+3}-y=\left(\sqrt{y^2-5y+3}-y\right)\cdot \frac{\sqrt{y^2-5y+3}+y}{\sqrt{y^2-5y+3}+y}$$

Note that your first cancellation is not correct because you have considered $\sqrt{x^2+5x+3}=-x$ which is not true whereas in the second case we have cancelled out $x^2-x^2$ which is a correct step.

  • 2
    $\begingroup$ I mean that's a nice hint but I can't see how that answers y question. There isn't much difference between $\lim_{x\to -\infty} x-x$ vs. $\lim_{y\to \infty}-y+y$ $\endgroup$
    – xotix
    Dec 3, 2018 at 14:35
  • $\begingroup$ @xotix I've added a specific observation about your wrong step. $\endgroup$
    – user
    Dec 3, 2018 at 14:43

Possibly the biggest problem in your first paragraph is that $\sqrt{\text{whatever}}$ is non-negative when we are working over the reals. Can you exhibit any choice of $x$ in the real numbers giving a negative value of $\sqrt{x^2 + 5 x + 3}$?


I will flesh out much of what would be considered a series of valid steps you may take in order to get something similar to the first step performed in the initial limit calculation. I will note that the second step that was taken is valid because we perform all of the algebra in the limit expression first, then we take a limit of what results.

The following line of reasoning shows that whenever you have $\sqrt{x^2+\beta x+\alpha}$ as a term (added or substracted, not multiplied or divided) in a limit, it may be replaced freely by $|x+\frac{\beta}{2}|$ inside the limit (of course slight adjustments must be made for it to be shown in general).

Note that for every $b>\frac{5}{2}$,

$$\lim_{x\to-\infty}\sqrt{x^2+2bx+b^2}+x \le\lim_{x\to-\infty}\sqrt{x^2+5x+3}+x.$$

To see why this is the case, here is a series of inequalities. First pick a $b>\frac{5}{2}$. Then pick $x$ so that $x\le\frac{3-b^2}{2b-5}$.

Then $$(2b-5)x\le 3-b^2,$$ $$2bx+b^2\le 5x+3,$$ $$x^2+2bx+b^2\le x^2+5x+3,$$ $$\sqrt{x^2+2bx+b^2}\le\sqrt{x^2+5x+3},$$ $$\sqrt{x^2+2bx+b^2}+x\le\sqrt{x^2+5x+3}+x,$$ $$\lim_{x\to-\infty}\sqrt{x^2+2bx+b^2}+x \le\lim_{x\to-\infty}\sqrt{x^2+5x+3}+x.$$

The intuition for this comes largely from working backward. So thus the claim is justified. So now we can say that

$$\lim_{b\to \frac{5}{2}^-}\lim_{x\to-\infty}\sqrt{x^2+2bx+b^2}+x \le\lim_{x\to-\infty}\sqrt{x^2+5x+3}+x,$$ $$\lim_{b\to \frac{5}{2}^-}|x+b|+x=\lim_{b\to \frac{5}{2}^-}-x-b+x=\lim_{b\to \frac{5}{2}^-}-b=-\frac{5}{2}\le\lim_{x\to-\infty}\sqrt{x^2+5x+3}+x.$$

Now since $3<\frac{25}{4}$, we have that.

$$\lim_{x\to-\infty}\sqrt{x^2+5x+3}+x\le \lim_{x\to-\infty}\sqrt{x^2+5x+\frac{25}{4}}+x=\lim_{x\to-\infty}|x+\frac{5}{2}|+x=\lim_{x\to-\infty}-x-\frac{5}{2}+x=\lim_{x\to-\infty}-\frac{5}{2}=-\frac{5}{2}.$$

So $$\lim_{x\to-\infty}\sqrt{x^2+5x+3}+x=-\frac{5}{2}.$$


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