In fact, if you plug $x=-2$ into the denominator of your last expression in the first formula, you get $\sqrt{\tfrac6{x^2}+\tfrac1x}-\tfrac2x=1+1=2$. So this would lead to limit equal to zero. The correct version would be, in my opinion, $-\sqrt{\tfrac6{x^2}+\tfrac1x}-\tfrac2x$ in the denominator. Notice that this leads to the indeterminate form $\tfrac00$, as expected. (We are using that $\sqrt{x^2}=|x|=-x$ for $x<0$. Since we are interested in the values close to $-2$, it suffices to work with negative $x$.) —
Martin Sleziak May 23 '16 at 13:33