Compute lim_{x->0} sin x / x.

Compute lim_{x->0} sin x / x.

• A. 0
• B. ∞
• C. 1
• D. -1

Answer: (C)

The main idea is how sine behaves for very small angles: sin x is essentially equal to x, so the ratio sin x / x should approach 1 as x goes to 0. A precise way to see this is with a squeeze. For x near 0 (and positive for the moment), we have sin x ≤ x ≤ tan x, and since tan x = sin x / cos x, this gives sin x ≤ x ≤ sin x / cos x. Rearranging to isolate sin x / x yields cos x ≤ sin x / x ≤ 1. As x → 0, cos x → 1, so by the squeeze theorem sin x / x → 1. The same reasoning works from the negative side, so the limit from both directions is 1. This is also echoed by the Taylor expansion sin x = x - x^3/6 + ..., so sin x / x = 1 - x^2/6 + ... → 1. Therefore, the value is 1.

The main idea is how sine behaves for very small angles: sin x is essentially equal to x, so the ratio sin x / x should approach 1 as x goes to 0. A precise way to see this is with a squeeze. For x near 0 (and positive for the moment), we have sin x ≤ x ≤ tan x, and since tan x = sin x / cos x, this gives sin x ≤ x ≤ sin x / cos x. Rearranging to isolate sin x / x yields cos x ≤ sin x / x ≤ 1. As x → 0, cos x → 1, so by the squeeze theorem sin x / x → 1. The same reasoning works from the negative side, so the limit from both directions is 1. This is also echoed by the Taylor expansion sin x = x - x^3/6 + ..., so sin x / x = 1 - x^2/6 + ... → 1. Therefore, the value is 1.