AP Calculus AB and BC

Limit of sin x / tan x as x Approaches 0 Is 1

The limit of sin x over tan x as x approaches 0 is 1. Simplifying first is the fastest route: dividing sine by tangent leaves cos x, which is continuous at 0 with value 1. No standard limit or L'Hopital is needed.

lim⁡x→0sin⁡xtan⁡x=1\lim_{x \to 0} \frac{\sin x}{\tan x} = 1

Settled by simplifying to cos x, then direct substitution.

Simplify before anything else

sin⁡xtan⁡x=sin⁡xsin⁡xcos⁡x=cos⁡x\frac{\sin x}{\tan x} = \frac{\sin x}{\frac{\sin x}{\cos x}} = \cos x

Once simplified the function is continuous at 00, so substitution gives cos⁡0=1\cos 0 = 1 immediately.

Simplify is a technique, not a preliminary

Direct substitution failing on the ORIGINAL form does not mean it will fail on an equivalent one. Rewriting is the first thing to try, ahead of every named limit technique.

The hole at the origin

The original expression is undefined at x=0x = 0 because tangent is 00 there, so the graph has a removable discontinuity: a hole at (0,1)(0, 1) on an otherwise ordinary cosine curve.

The mistakes students make

  • Reaching for L'Hopital before simplifying. It works, but it is more effort than one line of algebra.
  • Claiming the function IS cos⁡x\cos x. They agree everywhere except at the points where tangent is undefined or zero.

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

What is the limit of sin x / tan x at 0?

It is 11, because the expression simplifies to cos⁡x\cos x.

Is sin x / tan x the same as cos x?

Not quite. They agree except where tangent is zero or undefined, so the original has holes the cosine does not.