Answer
The limit of the given sequence is $0$.
Work Step by Step
Step 1:
Intuitively speaking, the given function is a limit of sine function and a polynomial function of degree $1$.
We know that sine function remains bounded throughout the domain while the polynomial function would increase for increasing values of $n$.
Thus, the denominator will dominate and the limit must be zero.
Step 2:
Let the given sequence be represented as ${c_n}=\frac{\sin^2 n}{n}$.
Consider two sequences as follows:
${a_n}=0$ and ${b_n}=\frac{1}{n}$.
Note that for each $n$, $0 \leq \frac{\sin^2 n}{n} \leq \frac{1}{n}$ holds. Conditions for Squeezing Theorem are satisfied.
The limit of both sequences, ${a_n}$ and ${b_n}$ is $0$. Thus, the limit of ${c_n}$ must also be $0$.
Result:
The limit of the given sequence is $0$.