Evaluate the limit as x approaches 0 of sin x / x.

Evaluate the limit as x approaches 0 of sin x / x.

• A. 1
• B. 0
• C. -1
• D. Infinity

Answer: (A)

The main idea is that near zero, sine behaves like its input when the input is measured in radians. So sin x / x should approach 1 as x → 0. A standard way to see this uses a geometric squeeze on the unit circle. For small positive x, sin x ≤ x ≤ tan x = sin x / cos x. Dividing by sin x (which is positive near 0) gives 1 ≤ x/ sin x ≤ 1/ cos x. Taking reciprocals yields cos x ≤ sin x/ x ≤ 1. As x → 0, both cos x and 1 approach 1, so by the Squeeze Theorem, sin x/ x → 1. Alternatively, you can view sin x as x minus a higher-order term, sin x = x − x^3/6 + ..., which gives sin x/ x = 1 − x^2/6 + ..., tending to 1. Thus the limit is 1. The other options would imply different behavior that doesn’t match the sine function’s local linearity in radians.

The main idea is that near zero, sine behaves like its input when the input is measured in radians. So sin x / x should approach 1 as x → 0.

A standard way to see this uses a geometric squeeze on the unit circle. For small positive x, sin x ≤ x ≤ tan x = sin x / cos x. Dividing by sin x (which is positive near 0) gives 1 ≤ x/ sin x ≤ 1/ cos x. Taking reciprocals yields cos x ≤ sin x/ x ≤ 1. As x → 0, both cos x and 1 approach 1, so by the Squeeze Theorem, sin x/ x → 1.

Alternatively, you can view sin x as x minus a higher-order term, sin x = x − x^3/6 + ..., which gives sin x/ x = 1 − x^2/6 + ..., tending to 1.

Thus the limit is 1. The other options would imply different behavior that doesn’t match the sine function’s local linearity in radians.