To evaluate the limit $\lim_{n\to\infty}\sin\left(\frac{\pi}{2}n\right),$ we need to analyze the behavior of the sequence $\sin\left(\frac{\pi}{2}n\right)$ as $n \to \infty$.

Step-by-Step Analysis:

1. Observe that $n$ is an integer, and the expression $\frac{\pi}{2}n$ takes on values such as $\frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi, \ldots$.

2. Consequently, the values of $\sin\left(\frac{\pi}{2}n\right)$ cycle through a repeating sequence: $\sin\left(\frac{\pi}{2}\right) = 1, \quad \sin(\pi) = 0, \quad \sin\left(\frac{3\pi}{2}\right) = -1, \quad \sin(2\pi) = 0.$

3. As $n \to \infty$, these values continue to repeat in the pattern: $1, 0, -1, 0, 1, 0, -1, 0, \ldots$.

Conclusion

Since the sequence does not approach a single value but instead oscillates between $1$, $0$, and $-1$, the limit does not exist.

$\lim_{n\to\infty}\sin\left(\frac{\pi}{2}n\right) \text{ does not exist.}$

Would you like further details or have any questions?

Related Questions:

1. What is an example of a sequence where the limit exists?
2. How do we determine the limit of trigonometric sequences in general?
3. What happens if we replace $\frac{\pi}{2}$ with other multiples of $\pi$?
4. Can we analyze similar limits with different periodic functions?
5. How does the concept of subsequential limits apply to this problem?

Tip:

For periodic functions, always check whether the argument leads to a repeating pattern when dealing with sequences.