Let $ n $ be a positive integer greater than $ 1 \cdot $
Find the following limits without using L'Hôpital's rule or series expansion :
$$ \left.1\right)\ \lim_{x\to 0}{\frac{1-\prod\limits_{k=1}^{n}{\cos{\left(kx\right)}}}{x^{2}}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.2\right)\ \ \ \ \ \ \ \ \ \ \lim_{x\to 0}{\frac{n!x^{n}-\prod\limits_{k=1}^{n}{\sin{\left(kx\right)}}}{x^{n+2}}} $$
$$ \left.3\right)\ \lim_{x\to 0}{\frac{1-\prod\limits_{k=1}^{n}{\cos^{k}{\left(kx\right)}}}{x^{2}}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.4\right)\ \lim_{x\to 0}{\frac{H_{2n}x^{2n^{2}+n}-\prod\limits_{k=1}^{2n}{\sin^{k}{\left(kx\right)}}}{x^{2n^{2}+n+2}}} $$
Where $ H_{n}=\frac{\left(n!\right)^{n}}{\prod\limits_{j=0}^{n-1}{j!}},\ \left(\forall n\in\mathbb{N}^{*}\right) \cdot $
The first one has given me $ \frac{n\left(n+1\right)\left(2n+1\right)}{12} $, the second one $ \frac{n\left(n+1\right)!\left(2n+1\right)}{36} $, the third one $ \frac{n^{2}\left(n+1\right)^{2}}{8} $, and the last one has given me $ \frac{H_{2n}n^{2}\left(2n+1\right)^{2}}{6} \cdot $
what's your approach to solve this problem ?
Here's a nice cheat inspired by deadly coronavirus outbreak boredom. I doubt a high school student knowing only that $\lim_{x\to 0}\frac{\sin x}{x}=1$ would be able to perform this, but this satisfies the conditions of the question and although it's effect is similar to L' Hopital's rule, it is very useful for addressing limits of a particular form. We shall proceed as Feynman dictates, with a derivative with respect to a parameter of the limit. Which parameter you say, since there's none? When there's no good parameter, just put one in there!
Assume the following limit exists:
$$L=\lim_{x\to 0}\frac{1-\prod_{k=1}^n f(kx)}{x^2}$$
where the function $f$ obeys $f(0)=1~,~ f(x)=f(-x)$ and is at least twice differentiable in the vicinity of the origin. It is easy to see that the first and second limits fall into this category (for the first obviously $f(x)=\cos x$ and for the second $f(x)=\text{sinc} ~x$).
Also consider $$\ell(a)=\lim_{x\to 0}\frac{1-\prod_{k=1}^n f(kax)}{x^2}=a^2L$$
This shows that $\ell(a)$ is a smooth function of the parameter $a$. Since this function is well behaved one can justify the interchange of a derivative and the limit and we can write since $f'(0)=0$:
$$\frac{d\ell}{da}=-\lim_{x\to 0}\prod_{k=1}^nf(kax)\sum_{k=1}^n\frac{k}{f(kax)}\frac{f'(kax)-f'(0)}{x}=-af''(0)\sum_{k=1}^nk^2=2La$$
whence we identify
$$L=-f''(0)\frac{n(n+1)(2n+1)}{12}$$
for the first limit $f''(0)=-1$, and for the second limit we observe that $L_2=n!L$, where L is the computed limit with $f(x)=\text{sinc}~x$ and $f''(0)=-\frac{1}{3}$, which confirms the results given.
We can play a similar game with the third and fourth limits which fall into the following category of limits with the same restrictions as above:
$$L=\lim_{x\to 0}\frac{1-\prod_{k=1}^n(f(kx))^k}{x^2}~~,~~ \ell(a)=\lim_{x\to 0}\frac{1-\prod_{k=1}^n(f(kax))^k}{x^2}=La^2$$ Taking the derivative again we compute: $$\frac{d\ell}{da}=-\lim_{x\to 0}\prod_{k=1}^n(f(kax))^k\sum_{k=1}^n\frac{k^2}{f(kax)}\frac{f'(kax)}{x}=-a\sum_{k=1}^nk^3f''(0)$$
and we conclude
$$L=-f''(0)\frac{n^2(n+1)^2}{8}$$
Again here we easily see that
$$L_3=\frac{n^2(n+1)^2}{8}$$
and $$L_4=\lim_{x\to 0}H_{2n}\frac{1-\prod_{k=1}^{2n}(\text{sinc}(kx))^k}{x^2}=H_{2n}\frac{n^2(2n+1)^2}{6}$$
as requested.