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Multiple Choice
Evaluate limn→−∞an and determine whether the sequence converges or diverges. an=300sin(nn2−120)
A
an=300, converges.
B
an=300, diverges.
C
an=DNE, converges.
D
an=DNE, diverges.
Verified step by step guidance
1
First, identify the sequence given: a_n = 300\sin\left(\frac{n^2 - 120}{n}\right).
To evaluate the limit as n approaches negative infinity, consider the expression inside the sine function: \frac{n^2 - 120}{n}.
Simplify \frac{n^2 - 120}{n} to n - \frac{120}{n}. As n approaches negative infinity, \frac{120}{n} approaches 0, so the expression simplifies to n.
The sine function, \sin(n), oscillates between -1 and 1 for all real numbers n. Therefore, \sin(n) does not settle to a single value as n approaches negative infinity.
Since \sin(n) oscillates indefinitely, the sequence a_n = 300\sin(n) does not converge to a single value, and thus the limit does not exist (DNE), indicating divergence.