AP Calculus AB and BC

Limit of sin x/(x + sin x) as x Approaches 0

The limit of sine x over the quantity x plus sine x as x approaches zero is one half. Dividing the numerator and the denominator by x turns every piece into the standard sine limit, and the answer becomes one over one plus one.

lim⁡x→0sin⁡xx+sin⁡x=12\lim_{x \to 0} \frac{\sin x}{x + \sin x} = \frac{1}{2}

Settled by dividing through by x to expose the standard limits.

Divide through by x

Substitution gives 0/00/0, so something has to be rearranged. Dividing every term by xx turns each piece into a quantity whose limit is already known.

sin⁡xx+sin⁡x=sin⁡xx1+sin⁡xx⟶11+1=12\frac{\sin x}{x + \sin x} = \frac{\dfrac{\sin x}{x}}{1 + \dfrac{\sin x}{x}} \longrightarrow \frac{1}{1+1} = \frac{1}{2}

Dividing by the dominant term is the same move that settles rational limits at infinity. Here the dominant term near zero is xx itself.

What makes the answer a half

Near zero, sin⁡x\sin x and xx are essentially the same size, so the denominator is essentially 2x2x while the numerator is essentially xx. Their ratio is 1/21/2.

That size comparison is worth internalising: it predicts the answer before any algebra, and it flags immediately if the algebra comes out wrong.

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

Would L'Hopital's rule work here?

Yes. Differentiating gives cos⁡x1+cos⁡x\frac{\cos x}{1 + \cos x}, which evaluates to 12\frac{1}{2} at zero. Dividing through is fewer steps and needs no form re-check.

What if the denominator were x minus sin x?

Completely different. Both leading terms cancel, leaving cubic behaviour, and the quotient runs to infinity rather than settling at a finite value.