Proving that $\displaystyle{\lim _{n\to\infty}} \frac{1}{n^2} = 0$ using limit arithmetic Let $n \in \mathbb{N}$ be the index of sequence $\{\frac{1}{n^2}\}_{n=1}^\infty$.
I'm assuming that:
$$\displaystyle{\lim _{n\to\infty}} \frac{1}{n^2} = \frac{\displaystyle{\lim _{n\to\infty}} 1}{\displaystyle{\lim _{n\to\infty}}n^2}, \displaystyle{\lim _{n\to\infty}}n^2 \neq 0$$
The problem is, after I prove that the limit of the numerator is 1, I cannot prove that the denominator has a limit because it's a fluttering/divergent sequence. Having no limit, and striving towards infinity, how can I possibly do limit arithmetic on them?
 A: The identity
$$\lim_{n\rightarrow\infty}\frac{a_n}{b_n}=\frac{\lim\limits_{n\rightarrow\infty}a_n}{\lim\limits_{n\rightarrow\infty}b_n}$$
does not hold when the limits on the RHS don't exist (or when the limit in the denominator would be $0$). The limit $\lim_{n\rightarrow\infty}n^2$ does not exist as the sequence is divergent, so you can not use the above identity. A valid way way to do limit arithmetic would be
$$\lim_{n\rightarrow\infty}\frac{1}{n^2}=\left(\lim_{n\rightarrow\infty}\frac{1}{n}\right)\cdot\left(\lim_{n\rightarrow\infty}\frac{1}{n}\right).$$
In this case, interchanging limit and product is possible, because the limits on the RHS each exist. They, of course, evaluate to $0$ and hence so does the original limit.
A: Just do delta-epsilon.  There's no trick.  
For any $\epsilon> 0$ let $N \ge \frac 1{\sqrt{\epsilon}}$.
If $n > N$ then $|\frac 1{n^2} - 0| =\frac 1{n^2} < \frac 1{N^2} = \epsilon$.
So $\lim_{n\to \infty}\frac 1{n^2} = 0$.
That's it.
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In general, you should have one time or another have been presented with a proof that 

if $\lim f(n) = +\infty$ then $\lim \frac 1{f(n)} = 0$.

The proof would be:  As $\lim_{n\to a|\infty} f(n)=+\infty$ then for any $N$ there is a condition for $n$ so that $f(n) > N$.  [If $\lim_{n\to a} f(n) = \infty$ the condition is there is a $\delta$ so tha $|n-a| < \delta$ will mean $f(n) >N$.  If $\lim_{n\to \infty} f(n) = \infty$ the condition is there is a $M$ so that $n > M$ will mean $f(n) > N$].  
For any $\epsilon > 0$ let $N = \frac 1{\epsilon}$ then if $n$ has the condition we know $f(n) > N$ so $\frac 1{f(n)} < \epsilon$.  So $\lim \frac 1{f(n)} = 0$.
....
So if we assume $\lim_{n\to \infty} n^2 = \infty$ (which is either very clear; or provable as for any $N>0$ if $n > \sqrt{N} \implies n^2 > N$) we know the $\lim_{n\to \infty}\frac 1{n^2} = 0$.
This is why we informally say $\frac 1\infty = 0$.  But that is informal and too many people say it without knowing what it means.
