Consider the function \[f\left( x \right)\] given by the double limit \[f\left( x \right)=\underset{n\to \infty }{\mathop{\lim }}\,\underset{t\to 0}{\mathop{\lim }}\,\dfrac{{{\sin }^{2}}\left( n!\pi x \right)}{{{\sin }^{2}}\left( n!\pi x+{{t}^{2}} \right)}\]; where \[x\] is irrational number.
(a) \[f\left( x \right)=0\]
(b) \[f\left( x \right)=1\]
(c) \[f\left( x \right)\] is undefined
(d) None

Hint: Since the second limit is \[n\to \infty \] and the first limit is \[t\to 0\]. The order of giving limits should be from left to right. Here we solve for limits in that order only. Also we need to check the limit from both neighbourhoods of ‘t’ that is we need to check for both \[t\to {{0}^{+}}\] and \[t\to {{0}^{-}}\], but \[n\to \infty \] have only one neighbourhood. If both the values are equal then only we can say that a given function is defined and equals to that value.

Complete step-by-step solution
Now let us consider the given function
\[f\left( x \right)=\underset{n\to \infty }{\mathop{\lim }}\,\underset{t\to 0}{\mathop{\lim }}\,\dfrac{{{\sin }^{2}}\left( n!\pi x \right)}{{{\sin }^{2}}\left( n!\pi x+{{t}^{2}} \right)}\]
Now let us solve this equation with the left neighbourhood of ‘t’ as follow \[\begin{align}
  & \Rightarrow f\left( x \right)=\underset{n\to \infty }{\mathop{\lim }}\,\underset{t\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{{{\sin }^{2}}\left( n!\pi x \right)}{{{\sin }^{2}}\left( n!\pi x+{{t}^{2}} \right)} \\
 & \Rightarrow f\left( x \right)=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{{{\sin }^{2}}\left( n!\pi x \right)}{{{\sin }^{2}}\left( n!\pi x+{{\left( 0 \right)}^{2}} \right)} \\
 & \Rightarrow f\left( x \right)=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{{{\sin }^{2}}\left( n!\pi x \right)}{{{\sin }^{2}}\left( n!\pi x \right)} \\
\end{align}\]
Here, after applying the first limit we got the same value in the numerator and denominator.
Whether \[x\] is irrational number or not it doesn’t matter if it has no effect on the limit
Now by applying the second limit we get
\[\begin{align}
  & \Rightarrow f\left( x \right)=\underset{n\to \infty }{\mathop{\lim }}\,\left( 1 \right) \\
 & \Rightarrow f\left( x \right)=1 \\
\end{align}\]
This is a finite and defined value.
Now let us calculate the right neighbourhood of ‘t’ as follows
\[\begin{align}
  & \Rightarrow f\left( x \right)=\underset{n\to \infty }{\mathop{\lim }}\,\underset{t\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{{{\sin }^{2}}\left( n!\pi x \right)}{{{\sin }^{2}}\left( n!\pi x+{{t}^{2}} \right)} \\
 & \Rightarrow f\left( x \right)=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{{{\sin }^{2}}\left( n!\pi x \right)}{{{\sin }^{2}}\left( n!\pi x+{{\left( 0 \right)}^{2}} \right)} \\
 & \Rightarrow f\left( x \right)=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{{{\sin }^{2}}\left( n!\pi x \right)}{{{\sin }^{2}}\left( n!\pi x \right)} \\
\end{align}\]
Here, after applying the first limit we got the same value in the numerator and denominator.
Whether\[x\] is irrational number or not it doesn’t matter if it has no effect on the limit
Now by applying the second limit we get
\[\begin{align}
  & \Rightarrow f\left( x \right)=\underset{n\to \infty }{\mathop{\lim }}\,\left( 1 \right) \\
 & \Rightarrow f\left( x \right)=1 \\
\end{align}\]
This is a finite and defined value.
Since the left neighborhood and right neighborhood of ‘t’ are finite and equal we can say that
\[f\left( x \right)\] is defined and equals to ‘1’.
So, option (b) is correct.

Note: Some students may conclude that the given function is undefined because the domain of \[\sin \theta \] is \[\theta \in R\]. But as \[x\] is an irrational number students conclude that the given function is not defined. We should not conclude things in the first place. After solving all the limits even if it happens then only we should say that function is undefined.