Compute the limit as x approaches 0 of sin x divided by x.

Compute the limit as x approaches 0 of sin x divided by x.

• A. 0
• B. 2
• C. 1
• D. ∞

Answer: (C)

When x is very close to zero, the sine function behaves like its input, so the ratio sin x over x should stay near 1. A clear way to see this uses a squeeze argument from the unit circle: for small positive x, sin x ≤ x ≤ tan x. Dividing the middle by x and using tan x = sin x / cos x gives sin x / x ≤ 1 and sin x / x ≥ cos x. So near zero, cos x ≤ sin x / x ≤ 1. Because cos x approaches 1 as x → 0, the entire expression is trapped between two curves that both approach 1, forcing the limit to be 1. An alternative view is through the series expansion sin x = x - x^3/6 + ..., which makes sin x / x = 1 - x^2/6 + ..., clearly tending to 1 as x → 0.

When x is very close to zero, the sine function behaves like its input, so the ratio sin x over x should stay near 1. A clear way to see this uses a squeeze argument from the unit circle: for small positive x, sin x ≤ x ≤ tan x. Dividing the middle by x and using tan x = sin x / cos x gives sin x / x ≤ 1 and sin x / x ≥ cos x. So near zero, cos x ≤ sin x / x ≤ 1. Because cos x approaches 1 as x → 0, the entire expression is trapped between two curves that both approach 1, forcing the limit to be 1.

An alternative view is through the series expansion sin x = x - x^3/6 + ..., which makes sin x / x = 1 - x^2/6 + ..., clearly tending to 1 as x → 0.