The squeeze theorem and trigonometric limits

Summary

This lesson obtains limits beyond the reach of the rules of the algebra of limits, by comparing the function with others whose limit is known. First it is proved that limits preserve non-strict inequalities; then comes the squeeze theorem, which reduces to the bound criterion of the preceding lesson without any new epsilon-delta argument, together with its most used case: the product of a bounded function and one that tends to zero. With the sine bound, proved in Algebra and Geometry II, it is shown that the sine and the cosine have a limit at every point, equal to their value. The inequality between the sine, the arc and the tangent then yields the fundamental limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$ , from which the rules already known deduce the other trigonometric limits. The problems apply the squeeze theorem, compute trigonometric limits by changes of variable and obtain the limit of $\frac{\arctan y}{y}$ .

Learning objectives

Prove the preservation of inequalities, by means of the preservation of sign, and, by means of the bound criterion, the squeeze theorem and the limit of the product of a bounded function and one that tends to zero.
Prove that the sine and the cosine have a limit at every point, equal to their value, and deduce that of the tangent.
Prove the fundamental limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$ and deduce from it the limits of $\frac{\tan x}{x}$ , $\frac{\sin kx}{x}$ , $\frac{1 - \cos x}{x^2}$ and $\frac{1 - \cos x}{x}$ .
Compute limits by comparison and trigonometric limits, justifying each step with the theorems of the lesson.

Limits beyond the reach of the rules

In the lesson on the algebra of limits, every $\frac{0}{0}$ form was resolved by transforming the expression until a common factor could be canceled. The quotient $\frac{\sin x}{x}$ , as $x$ tends to $0$ , does not admit that treatment: the sine is not a polynomial, and no factorization makes the factor $x$ appear in the numerator. Nor does the product rule decide the limit of $x \sin \frac{1}{x}$ at $0$ , because the factor $\sin \frac{1}{x}$ has no limit, as proposed problem 3 of the lesson on the limit of functions of a real variable asked you to prove.

Now, in both cases the function is trapped between two simpler ones: the absolute value of $x \sin \frac{1}{x}$ does not exceed $\abs{x}$ , and the quotient $\frac{\sin x}{x}$ lies, as will be seen, between $\cos x$ and $1$ . If this is so, it is worth proving once and for all that a function trapped between two functions that tend to the same number also tends to that number. The proofs of this lesson follow the plan of the preceding one: each result reduces to an inequality, and the bound criterion, or the tool deduced from it, does the rest.

The conventions of the preceding lesson remain in force: $f$ , $g$ and $h$ are real functions defined on the same set $A$ , the point $a$ is an accumulation point of $A$ , and saying that something holds “near $a$ ” means that there exists an $r > 0$ such that every $x$ in $A$ with $0 < \abs{x - a} < r$ satisfies it. Recall the tool with which the section on the three lemmas ended: if $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$ , every function $F$ that satisfies near $a$ a bound like the one below, with $C > 0$ , tends to $K$ .

\abs{F(x) - K} \leq C\left(\abs{f(x) - L} + \abs{g(x) - M}\right)