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? 

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 $\lim_{x \to \infty} \frac{2x^2 +1}{3x^2 + 4}$
