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

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