1.3 · The Squeeze Theorem and connecting representations

When two bounds force one limit

Suppose $\displaystyle g(x)\leq f(x)\leq h(x)$ for all x sufficiently close to a, except possibly at a itself. If both outer functions approach the same finite value L as x approaches a, the middle function must also approach L. This is the Squeeze Theorem. You do not need to know f(a).

All three conditions matter: the inequality must hold throughout a deleted neighborhood, both bounding limits must exist, and those limits must agree. Bounds approaching different numbers do not determine a unique limit.

$$g(x)\leq f(x)\leq h(x),\quad \lim_{x\to a}g(x)=\lim_{x\to a}h(x)=L$$

An oscillation can shrink to zero

The function $\displaystyle \sin(1/x)$ oscillates endlessly as x approaches 0 and has no limit there. However, $\displaystyle -1\leq\sin(1/x)\leq1$. Multiplying by the nonnegative number x² gives $\displaystyle -x^2\leq x^2\sin(1/x)\leq x^2$. Both bounds approach 0, so the product approaches 0.

Graphically, the oscillating curve stays inside the envelope $\displaystyle y=\pm x^2$. The oscillations never stop, but their height shrinks. Do not assume that a product has no limit just because one factor has no limit.

$$-x^2\leq x^2\sin(1/x)\leq x^2$$

Why sin(x)/x approaches 1

Measure angles in radians. For $\displaystyle 0<x<\frac{\pi}{2}$, a unit-circle comparison of a triangle, a sector, and a tangent triangle gives $\displaystyle \sin x<x<\tan x$. Dividing and rearranging gives $\displaystyle \cos x<\frac{\sin x}{x}<1$. Since $\displaystyle \cos x$ approaches 1, the Squeeze Theorem gives a right-hand limit of 1.

Because $\displaystyle \frac{\sin(-x)}{-x}=\frac{\sin x}{x}$, the ratio is even. The left-hand limit is also 1, so the two-sided limit is 1. The radians condition is essential: the sector-area comparison uses radian measure.

$$\lim_{x\to0}\frac{\sin x}{x}=1$$

Scale the input and rewrite the expression

To evaluate $\displaystyle \frac{\sin(5x)}{2x}$ near 0, rewrite it as $\displaystyle \frac{5}{2}\cdot\frac{\sin(5x)}{5x}$. The inner input 5x approaches 0, so the standard ratio approaches 1 and the result is $\displaystyle \frac{5}{2}$. The argument inside sine must match the denominator in the standard ratio.

For $\displaystyle \frac{1-\cos x}{x}$, multiply by $\displaystyle \frac{1+\cos x}{1+\cos x}$ to obtain $\displaystyle \frac{\sin x}{x}\cdot\frac{\sin x}{1+\cos x}$. The first factor approaches 1 and the second approaches 0. Their product approaches 0. These arguments use algebra and known limits, so derivative-based rules are unnecessary.

$$\lim_{x\to0}\frac{\sin(kx)}{x}=k$$

Connect evidence with justification

A plot or a table can suggest a limit and help you choose a method, but finitely many samples cannot establish behavior at every nearby input. For an oscillating function, a graph can also hide rapid oscillations when its resolution is too coarse.

Translate a bound into a graphical envelope, identify what happens to the envelope, then write the matching analytic limits. For AP-style justification, state the inequality, evaluate both outer limits, name the Squeeze Theorem, and give the resulting limit.