Limits: dividing by the highest power does not work in this case I believe to solve expressions like: 
$$\lim_{x\to \infty} (2x^2 + 1)$$
we need to divide each term in the numerator and denominator by the highest power. In the previous case, the highest power is $x^2$, so we get
$$\lim_{x \to \infty} \left(2 + \frac{1}{x^2} \right) = 2+0=2$$
Now for: 
$$\lim_{n \to \infty} \frac{-1}{\sqrt{n} + 2}$$
The answer given is $\frac{-1}{\infty}=0$ 
On the other hand, what I did was 
$$\lim_{x \to \infty} \frac{-1}{1 + \frac{2}{\sqrt{n}}}=\frac{-1}{1}=-1$$
What's wrong? 
 A: The rule you are referring to in your title "divide by the highest power" means that you divide every term in the given "rational expression" by  $x^k$ where $k$ is the highest power of $x$ in the entire limit expression. This includes terms in both the numerator and the denominator. If no denominator is explicitly shown, then the denominator is 1. 
For example, you
could apply this rule to
$$\lim_{x\rightarrow\infty} (x^2+1)=\lim_{x\rightarrow\infty} {x^2+1\over 1}
$$ 
to obtain 
$$\tag{0}\lim_{x\rightarrow\infty} {x^2+1\over1}
=\lim_{x\rightarrow\infty}  {1+{1\over x^2}\over{ 1\over x^2}}
= \lim_{x\rightarrow\infty}x^2 (1+\textstyle{1\over x^2} ). 
$$ 
The important observation is:
Note that you are just algebraically manipulating the limit expression; but, you cannot change the limit expression, as you did when you replaced $x^2+1$ by non-equivalent expression $1+{1\over x^2}$.  You can however replace  $x^2+1$ by the equivalent expression (as far as taking limits is concerned) $x^2(1+{1\over x^2})$, as was done above. Dividing  everything in the limit expression in the left hand side of $(0)$ by $x^2$ gives an equivalent expression because this amounts to multiplying both top and bottom by the same quantity (namely $1/x^2$).
I will also caution here that this rule is applicable in the specific scenario where 
$\ \ $1) you are taking a limit at infinity or minus ininifty 
and 
$\ \ $2) the limit expression is of the form $P(x)/Q(x)$ where both $p$ and $Q$ are linear combinations of powers of $x$ and the largest power of $x$ appearing in the entire expression is nonnegative.
In this case, the rule says to: find the highest power of $x$ in the entire expression and then divide both $P(x)$ and $Q(x)$ by $x^k$.  Note that this will lead to an equivalent expression since it amounts to multiplying bot top and bottom by the same quantity. Moreover, after doing this and simplifying, the limit will be easily determined.

Let's pause a second. There really is no reason to use this "divide by the highest power rule" when evaluating
$$
\lim_{x\rightarrow\infty} (x^2+1),
$$
as this limit is clearly infinite.
However, if you had to find
$$\tag{1}
\lim_{x\rightarrow\infty} (x^2-x^3)
$$
then things aren't so clear.  Here, you could use the "divide by the highest power   rule" (and note then, there is a 1 downstairs); but I think it's more clear to instead (and equivalently) factor the highest power of $x$
in the entire expression from both top and bottom (if there is one) and cancel (if you can). 
So for $(1)$, we would write:
$$ 
\lim_{x\rightarrow\infty} (x^2-x^3)=
\lim_{x\rightarrow\infty} x^3 (\textstyle{1\over x}-1)
$$
And now we can see that the limit is $-\infty$.
Or, consider the following example:
$$ \tag{2}
\lim_{x\rightarrow\infty} {2x^3-x+2\over 3x^3-2x+1}=
\lim_{x\rightarrow\infty} {x^3(2-{1\over x^2}+{2\over x^3})\over x^3( 3-{2\over x^2}+{1\over x^3})}=
\lim_{x\rightarrow\infty} { 2-{1\over x^2}+{2\over x^3} \over 3-{2\over x^2}+{1\over x^3} }
={2\over3}.
$$
Note that the last limit in $(2)$ could have been obtained by "dividing everything by $x^3$"; but it's of utmost importance that you divide all terms in both the numerator and denominator by $x^3$. Again, you can't change the limit expression.
For your second problem, dividing by $\sqrt n$ means you have to divide the $1$ in the numerator  by $\sqrt n$ too
$$
\lim_{n\rightarrow\infty}{-1\over \sqrt n +2}
=\lim_{n\rightarrow\infty}{-1/\sqrt n\over 1 +{2\over\sqrt n}}
={0\over 1}=0.
$$
As with the first example, though, this is a bit much for this limit.  The limit is clearly 0...
As mentiond by Andy, the method is better suited for something like:
$$\tag{3}
\lim_{n\rightarrow\infty}{-n^2\over \sqrt n +n^3}.
$$
Using the "divide trick", you would divide everything by $n^3$:
$$ 
\lim_{n\rightarrow\infty}{-n^2\over \sqrt n +n^3}
=\lim_{n\rightarrow\infty}{-{n^2\over n^3}\over {\sqrt n\over n^3} +1}
=\lim_{n\rightarrow\infty}{-{1\over n }\over {1\over n^{5/2}} +1}
={0\over 0+1}=0.
$$
Equivalently, use the "factor method" described above:
$$ 
\lim_{n\rightarrow\infty}{-n^2\over \sqrt n +n^3}
=\lim_{n\rightarrow\infty}{ n^3\cdot -{1\over n}\over n^3({1\over n^{5/2}}+1))}
=\lim_{n\rightarrow\infty}{-{1\over n }\over {1\over n^{5/2}} +1}
={0\over 0+1}=0.
$$
A: What you are saying in your calculations is $2x^2 +1 = (2 + 1/x^2)$, which is wrong. What you want is $2x^2 +1 = x^2(2 + 1/x^2)$, thus
$$\lim_{x \to \infty} 2x^2 +1 = \lim_{x \to \infty} x^2(2+ 1/x^2) = \lim_{x\to \infty} 2x^2 = \infty$$
for your second limit you are making the same mistake: $\sqrt{n} + 2 \neq 1+ 2/\sqrt{n}$, but $\sqrt{n} + 2 = \sqrt{n}(1+2/\sqrt{n})$. Thus,
$$\lim_{x \to \infty} \frac{-1}{\sqrt{n}+2} = \lim_{x \to \infty} \frac{-1}{\sqrt{n}(1+2/\sqrt{n})} = \lim_{x \to \infty} \frac{-1}{\sqrt{n}} = 0$$
This method is usually better suited for indeterminate cases, like $\lim_{x \to \infty} \frac{2x^2 +1}{3x^2 + 4}$
