AP Calculus AB and BC

Limit of sin^2(x)/x as x Approaches 0

The limit of sine squared x over x as x approaches zero is zero. Splitting the square into two factors leaves the standard sine limit, which tends to one, multiplied by a lone sine that tends to zero.

lim⁡x→0sin⁡2xx=0\lim_{x \to 0} \frac{\sin^{2} x}{x} = 0

Settled by splitting off the special trigonometric limit.

Split the square

sin⁡2xx=sin⁡x⋅sin⁡xx⟶0⋅1=0\frac{\sin^{2}x}{x} = \sin x \cdot \frac{\sin x}{x} \longrightarrow 0 \cdot 1 = 0

The product law applies because both factors have limits of their own. The second is the standard limit and equals 1; the first is just sin⁡x\sin x, which tends to 0.

The instinct to reach for the standard limit is right; the discipline is to peel off exactly one copy and treat whatever remains separately.

Sizes tell you the answer first

Near zero sin⁡x\sin x behaves like xx, so sin⁡2x\sin^{2}x behaves like x2x^{2} and the quotient behaves like x2x=x\frac{x^{2}}{x} = x, which vanishes.

That estimate also predicts the neighbouring cases: sin⁡2xx2→1\frac{\sin^{2}x}{x^{2}} \to 1 because the sizes match, and sin⁡2xx3\frac{\sin^{2}x}{x^{3}} is unbounded because the denominator wins.

Not sure which technique a limit wants?

The Limit Method Chooser walks the decision from direct substitution through factoring, the conjugate, and L'Hopital, and says why each one applies or fails.

Frequently asked questions

Why is the answer not 1?

Because only ONE factor of xx sits in the denominator against two factors of sine on top. Peeling off the standard limit leaves an extra sin⁡x\sin x, which drives the product to 0.

What is the limit of sin squared x over x squared?

It is 1. Both powers match, so the quotient is the square of the standard limit.