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

@Martin Sleziak, I didn't understand, why is there a negative sign there? — Mr Reality 1 hour ago

@MrReality You have $|x|=\sqrt{x^2}$. So for $x<0$ this gives you $x=-|x|=-\sqrt{x^2}$. This is explained in the comment above. — Martin Sleziak 1 hour ago

@Martin Sleziak Yes, I saw the earlier comment. I understand the reason for a minus sign in the examole you gave now but in your answer the term inside the square root is not $x^2$ here, it is $\sqrt{\tfrac6{x^2}+\tfrac1x}$. So why is a minus sign there? — Mr Reality 51 mins ago

@MrReality Maybe this is what you're after: $\frac{\sqrt{6+x}}{x}= \frac{\sqrt{6+x}}{-\sqrt{x^2}}= -\frac{\sqrt{6+x}}{\sqrt{x^2}}= -\sqrt{\frac{6+x}{x^2}}$. In any case, we're leaving too many irrelevant comments here and additionally each of the adds ping to the answerer. So perhaps it would be better to continue in chat if further comments are needed. — Martin Sleziak 18 secs ago