AP Calculus AB and BC

Limit of sin(1/x) at Infinity Is 0

The limit of sine of 1 over x as x approaches infinity is 0. As x grows, 1 over x tends to 0, and sine is continuous at 0, so the whole expression tends to sine of 0. The same formula has no limit as x approaches 0, where 1 over x runs off to infinity instead.

lim⁡x→∞sin⁡(1x)=0\lim_{x \to \infty} \sin\left(\frac{1}{x}\right) = 0

Settled by continuity of sine at 0.

Follow the inside first

Composite limits are settled from the inside out, provided the outer function is continuous at the value the inside is heading for. Here the inside is 1x\frac{1}{x}, which tends to 00, and sine is continuous at 00.

lim⁡x→∞sin⁡(1x)=sin⁡(lim⁡x→∞1x)=sin⁡0=0\lim_{x \to \infty}\sin\left(\frac{1}{x}\right) = \sin\left(\lim_{x \to \infty}\frac{1}{x}\right) = \sin 0 = 0

At x=1000x = 1000 the value is sin⁡(0.001)\sin(0.001), which is about 0.0010.001. The output is being squeezed toward 00 by its own argument.

Contrast this with x approaching 0

Send xx to 00 instead and 1x\frac{1}{x} grows without bound, so the sine runs through full cycles forever. Every output between −1-1 and 11 is hit infinitely often in any interval around the origin.

sin⁡(1x)=1 at every x=2(4k+1)π, k∈Z\sin\left(\frac{1}{x}\right) = 1 \text{ at every } x = \frac{2}{(4k+1)\pi}, \ k \in \mathbb{Z}

Those points crowd toward 00, and so do the points where the value is −1-1. That is an oscillating discontinuity, and the limit at 00 does not exist.

One formula, two answers

The behaviour of sin(1/x) depends entirely on where x is heading. Toward infinity the argument dies away and the limit is 0. Toward 0 the argument explodes and there is no limit at all.

The mistakes students make

Two of these come from importing the wrong picture, one from misreading the notation.

  • Answering that the limit does not exist, using the oscillation near x=0x = 0. Far out along the axis the argument is tiny and the oscillation has stopped.
  • Answering 11 by confusing this with lim⁡x→∞xsin⁡(1x)\lim_{x \to \infty} x\sin\left(\frac{1}{x}\right), which is the one that equals 11.
  • Reading sin⁡(1x)\sin\left(\frac{1}{x}\right) as 1sin⁡x\frac{1}{\sin x}. That is csc⁡x\csc x, which oscillates without bound and has no limit at infinity.

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(1/x) as x goes to infinity?

It is 00.

Why does sin(1/x) have no limit as x approaches 0?

Because 1x\frac{1}{x} becomes unbounded, so the sine cycles between −1-1 and 11 infinitely often near the origin and never settles.

Do I need the squeeze theorem for this?

No. Continuity of sine is enough here. The squeeze theorem is the tool for xsin⁡(1x)x\sin\left(\frac{1}{x}\right) as x→0x \to 0, where the oscillation has to be damped by a factor.