Math Problem Statement

Calculate the limits of each of the following sequences. If a sequence converges, enter its limit. If a sequence diverges, enter the letter X.

a. {sin(5n)/n} b. {sin(4n)/n} c. {sin(π(6n+1)/2) d. {nsin(6πn)}

Solution

Let's analyze each sequence one by one:

(a) {sin(5n)n}\left\{ \frac{\sin(5n)}{n} \right\}

  • The function sin(5n)\sin(5n) is bounded, meaning that for any nn, 1sin(5n)1-1 \leq \sin(5n) \leq 1.
  • As nn \to \infty, the denominator nn grows without bound while sin(5n)\sin(5n) oscillates between -1 and 1.
  • Therefore, sin(5n)n0\frac{\sin(5n)}{n} \to 0 as nn \to \infty.

Limit: 0

(b) {sin(4n)n}\left\{ \frac{\sin(4n)}{n} \right\}

  • Similarly, sin(4n)\sin(4n) is bounded by -1 and 1.
  • As nn \to \infty, the denominator nn increases without bound while sin(4n)\sin(4n) oscillates.
  • Hence, sin(4n)n0\frac{\sin(4n)}{n} \to 0 as nn \to \infty.

Limit: 0

(c) {sin(π(6n+1)2)}\left\{ \sin\left( \frac{\pi(6n + 1)}{2} \right) \right\}

  • First, simplify the expression inside the sine function: sin(π(6n+1)2)=sin(3nπ+π2)\sin\left( \frac{\pi(6n + 1)}{2} \right) = \sin\left( 3n\pi + \frac{\pi}{2} \right) Using the periodicity of sine, sin(x+2kπ)=sin(x)\sin(x + 2k\pi) = \sin(x), we get: sin(3nπ+π2)=sin(π2)=1\sin\left( 3n\pi + \frac{\pi}{2} \right) = \sin\left( \frac{\pi}{2} \right) = 1
  • Therefore, this sequence simplifies to 11 for all nn.

Limit: 1

(d) {nsin(6πn)}\left\{ n \sin(6\pi n) \right\}

  • The term 6πn6\pi n is a multiple of 2π2\pi, and we know that sin(k2π)=0\sin(k \cdot 2\pi) = 0 for any integer kk.
  • Thus, sin(6πn)=0\sin(6\pi n) = 0 for all nn.

Therefore, nsin(6πn)=n0=0n \sin(6\pi n) = n \cdot 0 = 0.

Limit: 0


Summary of Limits:

  • (a) limnsin(5n)n=0\lim_{n \to \infty} \frac{\sin(5n)}{n} = 0
  • (b) limnsin(4n)n=0\lim_{n \to \infty} \frac{\sin(4n)}{n} = 0
  • (c) limnsin(π(6n+1)2)=1\lim_{n \to \infty} \sin\left( \frac{\pi(6n + 1)}{2} \right) = 1
  • (d) limnnsin(6πn)=0\lim_{n \to \infty} n \sin(6\pi n) = 0

Would you like more details on any specific sequence?

Relative Questions:

  1. What is the general method to calculate limits for trigonometric sequences?
  2. Why do sine functions oscillate, and how does this affect limits?
  3. How does periodicity of trigonometric functions impact the convergence of sequences?
  4. What are common types of divergent sequences in trigonometry?
  5. How does L'Hopital's Rule apply to limit problems involving trigonometric functions?

Tip:

Always check for bounded behavior in oscillating functions like sine or cosine to determine the convergence behavior of sequences.

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Math Problem Analysis

Mathematical Concepts

Limits of Sequences
Trigonometric Functions
Oscillating Functions

Formulas

lim (n→∞) sin(kn)/n = 0 (for bounded sin(kn))
lim (n→∞) nsin(kπn) = 0 (for kπn multiples of 2π)
Sine periodicity: sin(x + 2πk) = sin(x)

Theorems

Boundedness of Sine and Cosine
Squeeze Theorem (for oscillating functions)
Periodicity of Trigonometric Functions

Suitable Grade Level

Grades 11-12