Chapter 1.2, Problem 9E

To determine

To find: The value of limit.

Answer to Problem 9E

The limit is 0.

Explanation of Solution

Given information: The limit is $\underset{x\to \infty }{\lim }\frac{1-\cos x}{x^2}$

Calculation:

The given expression is:

  $\underset{x\to \infty }{\lim }\frac{1-\cos x}{x^2}$

The range of $\cos x=(-1,1)$ :

  $-1\le \cos x\le 1$

Multiply by -1. Keep in mind that must invert the inequality signs when multiplying by a negative integer.

  $1\ge -\cos x\ge -1$

Rewrite with less than signs when using the sandwich theorem for the benefit of your sanity.

  $-1\le -\cos x\le 1$

Add 1:

  $0\le 1-\cos x\le 2$

Divide by $x^2$ , that is positive, without having to worry about the inequality signals being reversed.

  $0\le \frac{1-\cos x}{x^2}\le \frac{2}{x^2}$

Identify the "slices of bread boundaries’’.

  $\underset{x\to \infty }{\lim }0=0$

  $\begin{array}{c}\underset{x\to \infty }{\lim }\frac{2}{x^2}=\underset{x\to \infty }{\lim }2\cdot \underset{x\to \infty }{\lim }\frac{1}{x}\cdot \underset{x\to \infty }{\lim }\frac{1}{x}\\ =2\cdot 0\cdot 0\\ =0\end{array}$

The sandwich theorem holds true because the limits of the two "breads" are equal, and the limit of the "meat" is also equal.

  $\begin{array}{c}\underset{x\to \infty }{\lim }0\\ \Rightarrow \underset{x\to \infty }{\lim }\frac{2}{x^2}=0\\ \Rightarrow \underset{x\to \infty }{\lim }\frac{1-\cos x}{x^2}=0\end{array}$

Therefore the required limit is 0.