Dgdrxrt

Let $n in mathbbN$ be the index of sequence $frac1n^2_n=1^infty$.

I'm assuming that:
$$displaystylelim _ntoinfty frac1n^2 = fracdisplaystylelim _ntoinfty 1displaystylelim _ntoinftyn^2, displaystylelim _ntoinftyn^2 neq 0$$

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?

The identity
$$lim_nrightarrowinftyfraca_nb_n=fraclimlimits_nrightarrowinftya_nlimlimits_nrightarrowinftyb_n$$
does not hold when the limits on the RHS don't exist (or when the limit in the denominator would be $0$). The limit $lim_nrightarrowinftyn^2$ 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
$$lim_nrightarrowinftyfrac1n^2=left(lim_nrightarrowinftyfrac1nright)cdotleft(lim_nrightarrowinftyfrac1nright).$$
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.

Just do delta-epsilon. There's no trick.

For any $epsilon> 0$ let $N ge frac 1sqrtepsilon$.

If $n > N$ then $|frac 1n^2 - 0| =frac 1n^2 < frac 1N^2 = epsilon$.

So $lim_nto inftyfrac 1n^2 = 0$.

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 $lim frac 1f(n) = 0$.

The proof would be: As $lim_nto a f(n)=+infty$ then for any $N$ there is a condition for $n$ so that $f(n) > N$. [If $lim_nto a f(n) = infty$ the condition is there is a $delta$ so tha $|n-a| < delta$ will mean $f(n) >N$. If $lim_nto infty f(n) = infty$ the condition is there is a $M$ so that $n > M$ will mean $f(n) > N$].

For any $epsilon > 0$ let $N = frac 1epsilon$ then if $n$ has the condition we know $f(n) > N$ so $frac 1f(n) < epsilon$. So $lim frac 1f(n) = 0$.

....

So if we assume $lim_nto infty n^2 = infty$ (which is either very clear; or provable as for any $N>0$ if $n > sqrtN implies n^2 > N$) we know the $lim_nto inftyfrac 1n^2 = 0$.

This is why we informally say $frac 1infty = 0$. But that is informal and too many people say it without knowing what it means.