By factorization:

$$\lim_{x\to-\infty} \frac{\sqrt{x^2+2x}}{-x}\tag{1}$$

$$=\lim_{x\to-\infty} \frac{x\sqrt{1+\frac{2}{x}}}{-x}$$


If I input $x=-\infty$, the limiting value seems to be $-1$. But according to desmos, the limiting value should be $1$.

By L'Hopital's rule:

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

$$=\lim_{x\to-\infty} \frac{\dfrac{x+1}{\sqrt{x^2+2x}}}{-1}$$

$$=-\lim_{x\to-\infty} \dfrac{x+1}{\sqrt{x^2+2x}}$$

$$=-\lim_{x\to-\infty} \frac{1}{\dfrac{x+1}{\sqrt{x^2+2x}}}$$

$$=-\lim_{x\to-\infty} \dfrac{\sqrt{x^2+2x}}{x+1}$$

$$=-\lim_{x\to-\infty} \frac{\dfrac{x+1}{\sqrt{x^2+2x}}}{1}$$

$$=-\lim_{x\to-\infty} \dfrac{x+1}{\sqrt{x^2+2x}}$$

I can't get a determinate form.

My questions:

  1. How do I find $(1)$ using factorization?
  2. How do I find $(1)$ using L'Hopital's rule?


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    $\begingroup$ What is $\sqrt{x^2}$ ? $\endgroup$ May 16, 2022 at 6:17
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    $\begingroup$ A very simple way of finding the limit is to find the limit of its square, which is clearly $1$. Since the function is positive you can take square root at the end. $\endgroup$ May 16, 2022 at 6:19
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    $\begingroup$ It doesn't "work" in the sense that it doesn't resolve the limit, as you found by obtaining the same expression in your third and seventh lines. Eventually, you "give up" and apply factorization... $\endgroup$
    – user882145
    May 16, 2022 at 8:31
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    $\begingroup$ We are taking limit as $x \to -\infty$ and the denominator is positive for all $x <0$. So the given function is itself positive and the answer is the positive square root of $1$ which is $+1$. [It would have been $-1$ if the function was negative]. $\endgroup$ May 16, 2022 at 8:50
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    $\begingroup$ Yes, you surely have. $\endgroup$ May 16, 2022 at 8:56

4 Answers 4


Your approach is almost correct, you've just made a common mistake regarding square roots.

Writing out your manipulation of the numerator, you did $\sqrt{x^2 + 2x} = \sqrt{x^2(1 + \frac2x)} = \sqrt{x^2}\sqrt{1 + \frac2x} = x\sqrt{1 + \frac2x}.$ However, recall that because the principal square root is always positive (by definition) we actually have that $\sqrt{x^2} = |x|,$ and because we're looking at the limit as $x$ approaches $-\infty$ we consider negative $x,$ so $|x| = -x,$ explaining the sign discrepancy.

  • $\begingroup$ Understood. What about L'Hopital's rule? Does it not work here? $\endgroup$ May 16, 2022 at 8:29
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    $\begingroup$ @tryingtobeastoic As other answers/comments have indicated, L'Hopital's rule just doesn't yield a ton here because of how derivatives on radicals work, you kinda just get stuck in a loop. It still applies, the limits you got still go to $1,$ but it just doesn't help a ton here. (technically there is a little stunt you could pull to get that the limit must go to $1$ if it exists, but I really see no reason for it when it just ends up being a worse version of the argument that actually works) $\endgroup$ May 16, 2022 at 8:35

\begin{eqnarray} \frac{\sqrt{x^2+2x}}{-x}&=&\frac{\sqrt{x^2}\sqrt{1+\frac{2}{x}}}{-x}\\ &=&\frac{|x|\sqrt{1+\frac{2}{x}}}{-x}\\ &=&\frac{-x\sqrt{1+\frac{2}{x}}}{-x} \text{ since }x<0\\ &=&\sqrt{1+\frac{2}{x}} \end{eqnarray}


For the first question, the problem lies in the first step. It should be $$\lim_{x\to-\infty}\frac{\sqrt{x^2+2x}}{-x}$$ $$=\lim_{x\to-\infty}\frac{-x\sqrt{1+\frac{2}{x}}}{-x}$$ Notice that $\sqrt{x^2}$ is $|x|$, so not always $x$. In this case, for $x\to-\infty$, it should be $-x$.

For the second question, it seems to be not recommended to use L'Hopital's rule in this example.


This answer is based on @KaviRamaMurthy's comments.

Alternative way to find the limit:

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

$$\text{[$\sqrt{x^2+2x}$ is positive. Moreover, $-x$ is also positive. So, $\frac{\sqrt{x^2+2x}}{-x}$ is positive.]}$$

$$=\lim_{x\to-\infty} \frac{\sqrt{x^2+2x}}{|x|}$$

$$=\lim_{x\to-\infty} \frac{\sqrt{x^2+2x}}{\sqrt{x^2}}$$

$$=\sqrt{\lim_{x\to-\infty} \frac{x^2+2x}{x^2}}$$

$$\text{[Recall that $\lim_{x\to a} \sqrt[n]{f(x)}=\sqrt[n]{\lim_{x\to a} f(x)}$]}$$

$$=\sqrt{\lim_{x\to-\infty} 1+\frac{2}{x}}$$



Let us consider another case where the function is negative. Let us consider $\lim_{x\to-\infty} \frac{\sqrt{x^2+2x}}{x}$.

Another case:

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

$$=-\lim_{x\to-\infty} \frac{\sqrt{x^2+2x}}{-x}$$

$$=-\lim_{x\to-\infty} \frac{\sqrt{x^2+2x}}{|x|}$$

$$=-\lim_{x\to-\infty} \frac{\sqrt{x^2+2x}}{\sqrt{x^2}}$$

$$=-\sqrt{\lim_{x\to-\infty} \frac{x^2+2x}{x^2}}$$

$$=-\sqrt{\lim_{x\to-\infty} 1+\frac{2}{x}}$$



  • 1
    $\begingroup$ This is correct. $\endgroup$ May 16, 2022 at 9:59

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