<s>[INST] Find the limit of a multiplying term function when n tends to infinity. How do I evaluate the limit,
$$\lim_{n \to \infty}  \ \ \ \left(1-\frac1{2^2}\right)\left(1-\frac1{3^2}\right)\left(1-\frac1{4^2}\right)...\left(1-\frac1{n^2}\right)$$
I tried to break the $n^{\text{th}}$ term into $$\frac{(n+1)(n-1)}{n.n}$$ and then tried to do something but couldn't get anywhere. Please help as to how to solve such questions in general. Thanks.
 [/INST] The following will not be of help for such questions "in general", but it is too long for a comment.
Following Euler, we have
$$
\begin{align}
\frac{\sin x}{x} &=
\left( 1 - \frac{x}{\pi} \right) \left( 1 + \frac{x}{\pi} \right)
\left( 1 - \frac{x}{2 \pi} \right) \left( 1 + \frac{x}{2 \pi} \right)
\left( 1 - \frac{x}{3 \pi} \right) \left( 1 + \frac{x}{3 \pi} \right)
\dots \\
&=
\left[ 1 - \left( \frac{x}{\pi} \right)^2 \right]
\left[ 1 - \frac{1}{2^2} \left( \frac{x}{\pi} \right)^2 \right]
\left[ 1 - \frac{1}{3^2} \left( \frac{x}{\pi} \right)^2 \right]
\dots,
\end{align}
$$
or
$$
\frac{\sin x}{x} \cdot \frac{1}{1 - (x/\pi)^2} =
\left[ 1 - \frac{1}{2^2} \left( \frac{x}{\pi} \right)^2 \right]
\left[ 1 - \frac{1}{3^2} \left( \frac{x}{\pi} \right)^2 \right]
\dots.
$$
Hence the required product is
$$
\begin{align}
\lim_{x \to \pi} \frac{\sin x}{x} \cdot \frac{1}{1 - (x/\pi)^2}
&= \lim_{\epsilon \to 0} \frac{\sin \epsilon}{\pi - \epsilon} \cdot \frac{1}{1 - ([\pi - \epsilon]/\pi)^2} \\
&= \lim_{\epsilon \to 0} \frac{\epsilon}{\pi} \cdot \frac{1}{2 \pi \epsilon / \pi^2} \\
&= \frac{1}{2}.
\end{align}
$$
</s>