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Chapter 9: Problem 37

Find the limit of the following sequences or determine that the limit does not exist. Verify your result with a graphing utility. $$a_{n}=\frac{\sin (n \pi / 3)}{\sqrt{n}}$$

Short Answer

Expert verified
Answer: The limit of the sequence as $$n$$ approaches infinity is 0.

Step by step solution

01

Simplify the function if possible

In this case, the function is already simplified, so we don't need to simplify it further.
02

Analyze the behavior of the numerator and denominator separately

We observe that the numerator, which is $$\sin (n \pi / 3)$$, oscillates between -1 and 1, since the sine function has a maximum value of 1 and a minimum value of -1. The denominator $$\sqrt{n}$$ approaches infinity as $$n \to \infty$$.
03

Apply the Squeeze Theorem

Based on the behavior of the function observed in step 2, we can bound the function $$a_n$$ between two other functions which approach zero as $$n \to \infty$$. Since $$ -1 \leq \sin (n \pi / 3) \leq 1 $$, we can also write the inequality: $$-\frac{1}{\sqrt{n}}\leq \frac{\sin (n \pi / 3)}{\sqrt{n}} \leq \frac{1}{\sqrt{n}}$$ Now, as $$n \to \infty$$, we have: $$\lim_{n \to \infty} -\frac{1}{\sqrt{n}} = 0$$ and $$\lim_{n \to \infty} \frac{1}{\sqrt{n}} = 0 $$. Applying the Squeeze Theorem, we get: $$\lim_{n \to \infty} \frac{\sin (n \pi / 3)}{\sqrt{n}} = 0$$
04

Verify the result with a graphing utility

To verify the result, we can plot the given function $$a_n$$ using a graphing utility such as Desmos or GeoGebra. We will see that as $$n$$ increases, the values of $$a_n$$ get closer and closer to 0. This confirms our result that $$\lim_{n \to \infty} \frac{\sin (n \pi / 3)}{\sqrt{n}} = 0$$.

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