Can't prove the limit $$\lim_{n \to \infty}(n+2)^{2}\sin\frac{1}{n}=\infty.$$ by definition it should start: Let $M>0$. There exists an $N>0$ for every $n>N$: $$(n+2)^{2}\sin\frac{1}{n}>M.$$
Any hints about how to minimize the value up there ?
*I think It's not true to use $\sin x>-1$
I need to prove it by definition of limit
On
$$ \lim_{n\rightarrow \infty}(n+2)^2\sin\ \frac{1}{n} = \lim_{n\rightarrow \infty}(n+2)\frac{\sin\ \frac{1}{n}}{\frac{1}{n+2}} = \lim_{n\rightarrow \infty}(n+2) =\infty$$ since $$ \lim_{n\rightarrow \infty} \frac{\sin\ \frac{1}{n}}{ \frac{1}{n+2}} =1$$
On
Let us write it as
$$\lim_{x\rightarrow 0}(\frac{1}{x}+2)^{2}\sin[x]$$ which is the continuous version of the problem with $x=\frac{1}{n}$. We expand $\sin[x]$ by $$x-\frac{x^{3}}{6}+\frac{x^{5}}{120}\cdots$$ The left hand side is $2+\frac{1}{x^{2}}+2\frac{1}{x}$. We know $2\sin[x]$ and $2\frac{\sin[x]}{x}$ is bounded. But from the above expression it is clear that $\frac{\sin[x]}{x^{2}}$ is unbounded. The proof for the discrete case is the same.
I assume there are easier proofs like using L'Hospital rule, etc.
Well we need some technique here and its a fairly standard technique. We need to understand that if $x \in (0, \pi/2)$ then $$\dfrac{\sin x}{x} > \cos x = 1 - 2\sin^{2}(x/2) > 1 - 2(x/2)^{2} = 1 - \dfrac{x^{2}}{2}$$ Putting $x = 1/n$ we get $$n\sin(1/n) > 1 - \frac{1}{2n^{2}}$$ and hence $$(n + 2)^{2}\sin(1/n) > n^{2}\sin(1/n) > n\left(1 - \frac{1}{2n^{2}}\right) = n - \frac{1}{2n} > n - 1$$ Now I hope you can continue it from here.
Update: The inequalities I have started with are fundamental and provide a key step to the proof of $\lim_{x \to 0}\dfrac{\sin x}{x} = 1$. For the proof of such identities see this blog post (search for "Proof of Standard Limits" in the post). It is interesting to see how such simple inequalities can help in establishing limits only by using definition of limit.