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:

*

*How do I find $(1)$ using factorization?

*How do I find $(1)$ using L'Hopital's rule?


Related
 A: 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.
A: \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}
A: 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.
A: 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:
$$\lim_{x\to-\infty} \frac{\sqrt{x^2+2x}}{x}$$
$$=-\lim_{x\to-\infty} \frac{\sqrt{x^2+2x}}{-x}$$
$$=-\lim_{x\to-\infty} \frac{\sqrt{x^2+2x}}{|x|}$$
$$=-\lim_{x\to-\infty} \frac{\sqrt{x^2+2x}}{\sqrt{x^2}}$$
$$=-\sqrt{\lim_{x\to-\infty} \frac{x^2+2x}{x^2}}$$
$$=-\sqrt{\lim_{x\to-\infty} 1+\frac{2}{x}}$$
$$=-\sqrt{1}$$
$$=-1\text{(Ans.)}$$
