The Squeeze Theorem

Trapped between two curves

Does x^2\sin\!\left(\tfrac1x\right) have a limit as x\to 0? The sine factor swings between -1 and 1 faster and faster near 0. You cannot factor it, cancel it, or substitute. What else could pin the limit down?

The sine never leaves [-1,1], so the function is trapped between two simpler curves, -x^2 and x^2. Shrink the ochre window around a and watch the corridor the function is forced to stay inside.

Drag an edge of the ochre window, or use the controls. The indigo lines mark the lowest lower bound and the highest upper bound inside the window.

Limit

Window half-width

Show f Show the bounds Zoom with the window

Limit

Window

Lowest lower bound

Highest upper bound

Corridor height

The squeeze theorem

Suppose that for all x near a, except possibly at a itself,

g(x)\le f(x)\le h(x)\qquad\text{and}\qquad \lim_{x\to a}g(x)=\lim_{x\to a}h(x)=L.

Then \lim_{x\to a}f(x)=L.

Why should that be true? Pick any \varepsilon>0. Close enough to a, both g(x) and h(x) are within \varepsilon of L. Since f(x) sits between them, it is within \varepsilon of L as well. That is the corridor in the graph closing up.

Both conditions matter. Turn off "Show f" in the graph. Can you still say what the limit is? Now choose \sin(1/x). Its bounds do not meet. What can you conclude then?

The limit of sin x / x

Substituting gives \tfrac00, and there is nothing to factor. Comparing areas in the unit circle shows that for 0<|x|<\tfrac{\pi}{2},

\cos x\le\frac{\sin x}{x}\le 1.

Both bounds approach 1 as x\to 0, so the squeeze theorem gives \lim_{x\to 0}\frac{\sin x}{x}=1. This limit is the reason the derivative of \sin x is \cos x.

Solve it step by step

Each problem follows the same plan: bound the troublesome factor, build bounds for the whole function, find the limits of the bounds, and decide what the squeeze theorem says. The score counts steps you answer correctly on the first try.

Problems solved0

First-try steps0 of 0

Common mistakes

Multiplying an inequality by x: For x<0, multiplying -1\le\cos(1/x)\le 1 by x reverses the inequalities. Multiply by |x|, or bound the absolute value instead.
Bounds that do not meet: Knowing -1\le\sin(1/x)\le 1 traps the function in a corridor of height 2 that never closes. The squeeze theorem then says nothing.
Saying sin(1/x) approaches 0: As x\to 0, 1/x grows without bound, and \sin(1/x) keeps swinging between -1 and 1. It is the factor x^2 that drives x^2\sin(1/x) to 0.