Determining Limits Using the Squeeze Theorem

As $x \to 0$, the factor $\sin\frac{1}{x}$ swings between $-1$ and $1$ infinitely many times and never settles, so it has no limit, and the product law can't handle $x^2\sin\frac{1}{x}$. But the swinging factor can never make the product bigger than $x^2$ or smaller than $-x^2$, and both of those go to 0. A function pinned between two others that meet at the same height has nowhere else to go. That is the squeeze theorem, and it also proves the two trig limits behind every derivative of sine and cosine.

The squeeze theorem

The graph shows the idea. The solid curve wiggles without end near 0, but it lives in the narrowing gap between the dashed parabolas, and that gap closes to the single point $(0, 0)$.

Why it's true: pick any distance, however small. Close enough to $c$, both $g(x)$ and $h(x)$ are within that distance of $L$. Since $f(x)$ sits between them, it is too. That is exactly what $\lim\limits_{x \to c} f(x) = L$ means.

Each condition earns its place. The bounds must squeeze toward the same number: if $1 \le f(x) \le 3$ near $c$, you only learn that a limit, if one exists, lies between 1 and 3. For a two-sided limit, the inequality must hold close to $c$ on both sides; what happens far from $c$, or at $c$ itself, doesn't matter.

Building the bounds

The bounds usually come from the fact that sine and cosine never leave $[-1, 1]$, whatever their input:

Multiply through by a factor that is never negative, and the inequality survives. For $x \neq 0$, $x^2 > 0$, so

Both bounds approach 0 as $x \to 0$, so $\lim\limits_{x \to 0} x^2\sin\frac{1}{x} = 0$.

If the factor can be negative, multiplying flips the inequality wherever it is negative. The clean fix is to bound the absolute value: if $|f(x)| \le B(x)$ near $c$ and $B(x) \to 0$, then $-B(x) \le f(x) \le B(x)$, and the squeeze theorem gives $f(x) \to 0$. Example 1 uses this.

Why $\frac{\sin x}{x}$ approaches 1

Substituting $x = 0$ into $\frac{\sin x}{x}$ gives $\frac{0}{0}$, and no identity removes a common factor. The squeeze theorem settles it, using a picture of the unit circle.

Take $0 < \theta < \frac{\pi}{2}$, measured in radians. In the figure, three regions are nested: triangle $OPQ$ fits inside the sector $OPQ$, which fits inside triangle $OQT$. Their areas are:

• triangle $OPQ$: base $OQ = 1$, height $\sin\theta$, area $\frac{1}{2}\sin\theta$;
• sector $OPQ$: a fraction $\frac{\theta}{2\pi}$ of a circle of area $\pi$, so area $\frac{\theta}{2}$;
• triangle $OQT$: base $1$, height $QT = \tan\theta$, area $\frac{1}{2}\tan\theta$.

Nested regions have increasing areas, so $\frac{1}{2}\sin\theta < \frac{1}{2}\theta < \frac{1}{2}\tan\theta$, which gives

Divide through by $\sin\theta$, which is positive here: $1 < \frac{\theta}{\sin\theta} < \frac{1}{\cos\theta}$. Taking reciprocals of positive numbers reverses both inequalities:

As $\theta \to 0^+$, $\cos\theta \to 1$ and the constant 1 stays 1, so the squeeze theorem gives $\lim\limits_{\theta \to 0^+}\frac{\sin\theta}{\theta} = 1$. For negative $\theta$, $\frac{\sin(-\theta)}{-\theta} = \frac{-\sin\theta}{-\theta} = \frac{\sin\theta}{\theta}$, so the expression takes the same values on both sides of 0, and the left-hand limit is also 1.

The sector area $\frac{\theta}{2}$ is where radians matter. In degrees the sector's area is $\frac{\pi\theta}{360}$, and the same argument gives $\frac{\sin x^\circ}{x} \to \frac{\pi}{180}$ instead of 1. Calculus uses radians so that this limit is 1.

Why $\frac{1 - \cos x}{x}$ approaches 0

This one also starts as $\frac{0}{0}$. For $0 < x < \frac{\pi}{2}$, use $1 - \cos^2 x = \sin^2 x$:

Since $1 + \cos x > 1$ here, $0 < 1 - \cos x < \sin^2 x$. Divide by $x > 0$ and use $\frac{\sin x}{x} < 1$ from the previous section:

Both 0 and $\sin x$ approach 0, so $\lim\limits_{x \to 0^+}\frac{1 - \cos x}{x} = 0$. For negative $x$, replacing $x$ by $-x$ changes only the sign of the denominator, so the values on the left are the negatives of those on the right, and they approach $-0 = 0$ as well.

So arrange the expression to match before using either one. For example, $\frac{\sin 5x}{2x} = \frac{5}{2}\cdot\frac{\sin 5x}{5x} \to \frac{5}{2}$, and $\frac{\tan x}{x} = \frac{\sin x}{x}\cdot\frac{1}{\cos x} \to 1 \cdot 1 = 1$.

Worked examples

Where students lose points

On the exam

The two trig limits come inside quotients that you rearrange first, as in Example 3. The squeeze theorem itself is a justification: when a question gives you an inequality like the one in Example 4, a complete answer states the inequality, shows that both bounds have the same limit, and names the squeeze theorem. Topic 2.7 uses both trig limits to find the derivatives of $\sin x$ and $\cos x$.