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

Question

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?

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Related

Answer 1

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.

Answer 2

\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}

Answer 3

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.

Answer 4

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:

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

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

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

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

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

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

$$=-\sqrt{1}$$

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