# How to find $\lim_{x\to-\infty} \frac{\sqrt{x^2+2x}}{-x}$?

By factorization:

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

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

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

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?

Related

• What is $\sqrt{x^2}$ ? May 16, 2022 at 6:17
• 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. May 16, 2022 at 6:19
• 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...
– user882145
May 16, 2022 at 8:31
• 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]. May 16, 2022 at 8:50
• Yes, you surely have. May 16, 2022 at 8:56

## 4 Answers

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.

• Understood. What about L'Hopital's rule? Does it not work here? May 16, 2022 at 8:29
• @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) 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:

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

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

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

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

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

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

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

$$=\sqrt{1}$$

$$=1\text{(Ans.)}$$

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:

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

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

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

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

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

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

$$=-\sqrt{1}$$

$$=-1\text{(Ans.)}$$

• This is correct. May 16, 2022 at 9:59