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.

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
$$
\begin{align}
\lim_{x \to \pi} \frac{\sin x}{x} \cdot \frac{1}{1 - (x/\pi)^2}
&= \lim_{\epsilon \to 0} \frac{\sin \epsilon}{\pi - \epsilon} \cdot \frac{1}{1 - ([\pi - \epsilon]/\pi)^2} \\
&= \lim_{\epsilon \to 0} \frac{\epsilon}{\pi} \cdot \frac{1}{2 \pi \epsilon / \pi^2} \\
&= \frac{1}{2}.
\end{align}
$$
