<s>[INST] 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
 [/INST] 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.
</s>