Determining a limit I'm having troubles showing that
$$
\lim_{x\rightarrow -\infty} \frac{\sqrt{x^2+2x}}{x} = -1.
$$
In particular, why is the following derivation wrong?
$$
\lim_{x\rightarrow -\infty} \frac{\sqrt{x^2+2x}}{x} = \lim_{x\rightarrow -\infty} \frac{x\sqrt{1+2/x}}{x} =  \lim_{x\rightarrow -\infty} \sqrt{1+2/x} = \sqrt{1+\lim_{x\rightarrow -\infty}(2/x)} = 1.$$

Your mistake is here
$$
\lim_{x\rightarrow -\infty} \frac{\sqrt{x^2+2x}}{x} = \lim_{x\rightarrow -\infty} \frac{x\sqrt{1+2/x}}{x} = \ldots = 1.
$$
The correct is 
$$
\lim_{x\rightarrow -\infty} \frac{\sqrt{x^2+2x}}{x} = \lim_{x\rightarrow -\infty} \frac{-x\sqrt{1+2/x}}{x} = \ldots = -1
$$
since $x<0$ ($\sqrt{x^2}=|x|$, for example $\sqrt{(-1254)^2}=1254$).
