1.8 Determining Limits Using the Squeeze Theorem

Learning objectives

—LIM-1.E—Determine the limits of functions using equivalent expressions for the function or the squeeze theorem—Topic 1.8• LIM-1.E Determine the limits of functions using equivalent expressions for the function or the squeeze theorem.• LIM-1.E.2 The limit of a function may be found by using the squeeze theorem.• Enduring understanding LIM-1: Reasoning with definitions, theorems, and properties can be used to justify claims about limits.

Use the squeeze theorem when a function is difficult to evaluate directly but can be trapped between two simpler functions that approach the same value.

g(x)\le f(x)\le h(x)\text{ near }c,\qquad \lim_{x\to c}g(x)=\lim_{x\to c}h(x)=L\quad\Longrightarrow\quad\lim_{x\to c}f(x)=L

Find a lower and upper bound for the target function.
Verify that both inequalities hold for all sufficiently close inputs, possibly excluding $x=c$ .
Evaluate the two bounding limits.
Conclude the target limit only if both bounds approach the same value.

|\cos(1/x)|\le1;\Longrightarrow;|x\cos(1/x)|\le|x|;\Longrightarrow;-|x|\le x\cos(1/x)\le|x|.

As $x\to0$ , both $-|x|$ and $|x|$ approach $0$ . Therefore $\displaystyle\lim_{x\to0}x\cos(1/x)=0$ by the squeeze theorem, even though $\cos(1/x)$ keeps oscillating.

The bounds must work throughout a neighborhood, not merely at selected points, and their limits must agree. Using $-|x|$ and $|x|$ is essential here: multiplying $-1\le\cos(1/x)\le1$ directly by a negative $x$ would reverse the inequality signs.

Next Topic

—1.9 Connecting Multiple Representations of LimitsRead