$$\lim_{n\to\infty}\left(1-\frac{2}{3}\right)^{\tfrac{3}{n}}\cdot\left(1-\frac{2}{4}\right)^{\tfrac{4}{n}}\cdot\left(1-\frac{2}{5}\right)^{\tfrac{5}{n}}\cdots\left(1-\frac{2}{n+2}\right)^{\tfrac{n+2}{n}}$$
I mean,I tried to use Sandwitch rule but it didnt work.
Hint: Since $1+x\le e^x$, we have that $1-\frac{x}{1+x}\le e^{-\frac{x}{1+x}}\implies e^{\frac{x}{1+x}}\le1+x$. Therefore, $$ e^{-2/k}\le1-\frac2{k+2}\le e^{-2/(k+2)} $$ and, thus, $$ e^{-2\left(1+\frac2k\right)}\le\left(1-\frac2{k+2}\right)^{k+2}\le e^{-2} $$ Use this to bound the product. Then take $n^{\text{th}}$ roots and apply the Squeeze Theorem.
Hint: Rewrite your quantity as $$ \exp\left(\sum_{k=3}^{n+2}\frac{k}{n} \ln\left(1-\frac{k-1}{k}\right) \right) = \exp\left(\sum_{k=3}^{n+2}\frac{k}{n} \ln\left(1-\frac{2}{k}\right) \right) $$ and use Cesàro's theorem to find the limit of $\sum_{k=3}^{n+2}\frac{k}{n} \ln\left(1-\frac{2}{k}\right)$ before concluding by continuity of the exponential.
For Cesàro's theorem, observe that $a_k = k\ln(1-\frac{2}{k})\xrightarrow[k\to\infty]{} -2$.
The $m^{th}$ term is \begin{align} a_m & = \prod_{n=1}^m \left(1-\dfrac2{n+2}\right)^{(n+2)/m} = \prod_{n=1}^m \left(\dfrac{n}{n+2}\right)^{(n+2)/m} = \left(\dfrac{\prod_{n=1}^m n^{(n+2)/m}}{\prod_{n=1}^m (n+2)}\right)^{(n+2)/m}\\ & = \left(\dfrac{\prod_{n=1}^m n^{n/m}}{\prod_{n=3}^{m+2} n^{n/m}}\right) \prod_{n=1}^m n^{2/m} = \dfrac{2^{2/m}(m!)^{2/m}}{(m+1)^{(m+1)/m}(m+2)^{(m+2)/m}} \end{align} This gives us that \begin{align} \log(a_m) & = \dfrac{2\log2}m + \dfrac{2\log(m!)}m - \dfrac{m+1}m \log(m+1) - \dfrac{m+2}{m+1} \log(m+2)\\ & = \dfrac{2\log2}m + \dfrac{2}m\left(m\log(m)-m + \mathcal{O}(\log(m))\right) - \dfrac{m+1}m \log(m+1) - \dfrac{m+2}{m+1} \log(m+2)\\ \end{align} where the last step relies on DeMoivre's identity.
Finishing step $1$:
This gives us$$\log(a_m) = -2 + 2 \log(m) - \log(m+1)-\log(m+2) + \mathcal{O}\left(\dfrac{\log(m)}m \right)$$
Finishing step $2$:
Taking $m \to \infty$, we obtain$$\lim_{m \to \infty} \log(a_m) = -2$$
