Solve a seemingly simple limit $\lim_{n\to\infty}\left(\frac{n-2}n\right)^{n^2}$ $$\lim_{n\to\infty}\left(\frac{n-2}n\right)^\left(n^2\right)$$
Why does this go to 0? Why can I not just divide each item in the fraction by n and assume it would go to 1?
 A: A subtlety not fully appreciated by many beginning calculus students worldwide is that all occurences of the variable (here $n$) are tied together. In other words they go to infinity hand-in-hand. There are many occasions, where the limit processes can be separated, and done one at a time, but there are THEOREMS governing those situations. Yes, they look trivial, their proofs are usually not very difficult, but yet they are powerful! Because they tell that we can, in a way, treat some occurences of the variable $n$ as if the others did not exist.
For the purposes of explaining this difficulty I will study a different limit, namely
$$
\lim_{n\to\infty}\left(\frac{n-2}n\right)^n.
$$
Many a beginning student wants to conclude that the fraction tends to $1$, and thus the answer should also be $1$, because the limit of $1^n$ as $n\to\infty$ is (quite correctly!) equal to $1$. But effectively these students are calculating an iterated limit
$$
\lim_{m\to\infty}\left(\lim_{n\to\infty}\left(\frac{n-2}n\right)\right)^m,
$$
because they first study what happens to fraction $(n-2)/n$, and only then consider the effect of raising the result to a high power.
So the justification is exactly the one you proffered in your OP, and it is correct for this iterated limit. Observe that here first $n$, then $m$ tends to infinity. I indicated this by using parentheses to make it clear that $n\to\infty$ first (first working inside the parens).
But we might also approach this differently, and reverse the order of the two limit process, and study the following limit
$$
\lim_{n\to\infty}\left(\lim_{m\to\infty}\left(\frac{n-2}n\right)^m\right)
$$
instead.
Here a student could (again correctly) argue that w.l.o.g. we can assume that $n>2$. Therefore the fraction $(n-2)/n$ is between $0$ and $1$. If $q\in(0,1)$, then we know (perhaps a result derived in connection with a geometric series?) that 
$$
\lim_{m\to\infty}q^m=0.
$$
As this works for all $n>2$, the answer to this limit question should be $\lim_{n\to\infty}0=0$.
Again, this reasoning is correct, and applies to the limit process, where first $m$ and only then $n$ tends to infinity.
Yet we know from our studies revolving around the Napier's constant $e$ that the answer to the limit
$$
\lim_{n\to\infty}\left(\frac{n-2}n\right)^n=e^{-2},
$$
i.e. neither zero nor one but something in-between.
What went wrong? We forgot that the $n$ that we plug in when calculating the value of the fraction $(n-2)/n$ is the same as the exponent $n$. But why should we care, a student may cry? How is this any different from a calculation like
$$
\lim_{n\to\infty}\frac{n+3}{2n-7}\cdot\frac{3n+5}{n-4},
$$
where we CAN calculate it like
$$
\lim_{m\to\infty}\left(\lim_{n\to\infty}\frac{n+3}{2n-7}\cdot\frac{3m+5}{m-4}\right)=\left(\lim_{n\to\infty}\frac{n+3}{2n-7}\right)\cdot\left(\lim_{m\to\infty}\frac{3m+5}{m-4}\right)=\frac12\cdot\frac31=\frac32?
$$
Well, that's exactly because there is a theorem saying that when
$\lim_{n\to\infty}a_n=A$ and $\lim_{n\to\infty}b_n=B$, and BOTH $A$ and $B$ are real numbers, i.e. not infinities, then we also have $\lim_{n\to\infty}a_nb_n=AB$. That result is very simple. And there is a similar result for the powers: $\lim_{n\to\infty}a_n^{b_n}=A^B$ (provided that $A>0$). But this only applies to (finite) real numbers $A,B$. If you try to apply this result to the sequence
$$
\left(\frac{n-2}n\right)^n,
$$
you see that the premises do not hold, because here the exponents tend to infinity as opposed to some number $B$. Therefore we must look for a different approach.
This kind of subtleties are the reason, why the calculus textbooks devote a lot of time to the so called indeterminate (is that the correct English term?) cases such as $0/0$, $\infty/\infty$, $\infty\cdot0$, $1^\infty$, where the usual shortcuts cannot be justified by a suitable theorem (because no such theorem exists, and the answer always `depends').
A: $$\lim_{n\to\infty} \left(\frac{n-2}n\right)^{n^2}=\lim_{n\to\infty}e^{n^2\log(1-2/n)}=\lim_{n\to\infty}e^{n^2.(-2/n)}=\lim_{n\to\infty}e^{-2n}=0$$
A: $$\lim_{n\to\infty} \left(\frac{n-2}n\right)^{n^2} = \lim_{n\to\infty} \left(\underbrace{\left(1 - \frac 2 n\right)^n}_{\to e^{-2}}\right)^n = 0$$
Edit: Since what I'm doing does not seem kosher (see the comments), I'll explain: If the "inner" limit had been $1$, then indeed this argument fails. However, if it goes to something strictly between $0$ and $1$, then the argument is fine, and it arises from an implicit use of the sandwich rule. (After you do this sort of thing a lot, you tend to leave these details unwritten.)
Here is a more accurate solution:
First of all, we have
$$\lim_{n \to\infty} \left(1-\frac 2 n\right)^n = e^{-2}$$
Therefore, for large enough $n$, we have
$$\left(1-\frac 2 n\right)^n < c$$
where $c$ is any constant you like strictly between $e^{-2}$ and $1$. Therefore, we have (for large enough $n$),
$$0 \le \left( \frac {n-2} n \right) ^{n^2} < c^n \to 0$$
which by the sandwich rule implies the required result.
A: $$\lim_{n\to\infty}\left(\frac{n-2}n\right)^\left(n^2\right)$$
$$\lim_{n\to\infty}\left(1 -2/n\right)^{n^2}$$
$$e^{\lim_{n\to\infty}\left({\left(\frac{-2}{n}\right)}\times n^2\right)}$$
$$e^{-\infty} =0;$$
