Puzzled by $\lim\limits_{x \to - \infty} \sqrt{x^2+x}-x$

The difficulty here is in "factoring out" the $x$. What you really did there is this:

You claim that $x^2+x=x^2(1+\frac{1}{x})$, and that $\sqrt{x^2(1+\frac{1}{x})}=\sqrt{x^2}\sqrt{1+\frac{1}{x}}$. This is perfectly fine so far, as long as the latter square root is defined. The problem is in the last step: you've claimed, from here, that $\sqrt{x^2}=x$. And that's not true!

Remember: in general, $\sqrt{x^2}=\lvert x\rvert$, not $x$. And if $x<0$, then $\lvert x\rvert=-x$, not $x$.

So, it should be $$ \sqrt{x^2+x}=\sqrt{x^2}\sqrt{1+\frac{1}{x}}=\lvert x\rvert\sqrt{1+\frac{1}{x}}=-x\sqrt{1+\frac{1}{x}}. $$

If $x$ is negative, then the square root must be manipulated differently; that is, we have

$$\sqrt{x^2 + x} = \sqrt{x^2} \sqrt{1 + \frac 1 x} = |x| \sqrt{1 + \frac 1 x}$$

But since $x < 0$, $|x| = -x$. This leads us to taking the limit of

$$\frac{x}{-|x|\sqrt{1 + \frac 1 x} + x} = \frac{1}{-\sqrt{1 + \frac 1 x} + 1}$$

Letting $x \to -\infty$, the denominator tends to zero.

In addition to all the other answers, just note that if we set

$y = -x$ then we have

$$\lim_{y \to +\infty} \sqrt{y^2 - y} + y$$ which is clearly $+\infty$

Setting $\displaystyle-\frac1x=h\iff x=-\frac1h$

$$\lim_{x\to-\infty}\left(\sqrt{x^2+x}-x\right)=\lim_{h\to0^+}\left(\sqrt{\frac1{h^2}-\frac1h}+\frac1h\right)$$ $$=\lim_{h\to0^+}\frac{\sqrt{1-h}+1}h\text{ as }h>0$$

$$=\lim_{h\to0^+}\frac{(1-h)-1}{h(\sqrt{1-h}-1)} $$

$$=-\lim_{h\to0^+}\frac1{\sqrt{1-h}-1}\text{ as } h\ne0\text{ as }h\to0^+ $$

$$=-\frac1{1-1}$$