Determine lim x->0 sin(1/x).

Determine lim x->0 sin(1/x).

• A. 0
• B. 1
• C. DNE
• D. -1

Answer: (C)

When x gets arbitrarily close to zero, 1/x becomes unbounded and the sine of that rapidly changing argument keeps oscillating between -1 and 1 without settling on a single value. To see why the limit doesn’t exist, look at two sequences approaching zero that give different results: - Let x_n = 1/(2π n). Then 1/x_n = 2π n and sin(1/x_n) = sin(2π n) = 0 for all n, so this path tends to 0. - Let x_n = 1/(π/2 + 2π n). Then 1/x_n = π/2 + 2π n and sin(1/x_n) = sin(π/2 + 2π n) = 1 for all n, so this path tends to 1. Because approaching zero along these two paths yields different limits, the overall limit does not exist.

When x gets arbitrarily close to zero, 1/x becomes unbounded and the sine of that rapidly changing argument keeps oscillating between -1 and 1 without settling on a single value. To see why the limit doesn’t exist, look at two sequences approaching zero that give different results:

• Let x_n = 1/(2π n). Then 1/x_n = 2π n and sin(1/x_n) = sin(2π n) = 0 for all n, so this path tends to 0.

• Let x_n = 1/(π/2 + 2π n). Then 1/x_n = π/2 + 2π n and sin(1/x_n) = sin(π/2 + 2π n) = 1 for all n, so this path tends to 1.

Because approaching zero along these two paths yields different limits, the overall limit does not exist.