Chapter 2.2, Problem 25E

(a)

To determine

The value of $\underset{x\to 0}{\lim }\frac{\cos 2x-\cos x}{x^2}$ by graphing and find where it crosses the y-axis.

Answer to Problem 25E

The value of $\underset{x\to 0}{\lim }\frac{\cos 2x-\cos x}{x^2}=-1.5$.

Explanation of Solution

Using the online graphing calculator and draw the graph of the function $f\left(x\right)=\frac{\cos 2x-\cos x}{x^2}$ as shown below in Figure 1.

Zoom in the graph towards the point where the graph crosses the y-axis as shown below in Figure 2.

From Figure 2, it is observed that graph of $f\left(x\right)=\frac{\cos 2x-\cos x}{x^2}$ approaches −1.5 as x approaches 0 from either side.

That is, $\underset{x\to 0^{-}}{\lim }\frac{\cos 2x-\cos x}{x^2}=\underset{x\to 0^{+}}{\lim }\frac{\cos 2x-\cos x}{x^2}=-1.5$.

Since the right hand limits and the left hand limits are equal, the value of $\underset{x\to 0}{\lim }\frac{\cos 2x-\cos x}{x^2}$ exists.

Thus, the value of $\underset{x\to 0}{\lim }\frac{\cos 2x-\cos x}{x^2}=-1.5$.

(b)

To determine

The answer obtained in part (a) by evaluating the value of $\frac{\cos 2x-\cos x}{x^2}$ as x approaches 0.

Answer to Problem 25E

The value of $\underset{x\to 0}{\lim }\frac{\cos 2x-\cos x}{x^2}=-1.5$.

Explanation of Solution

As x approaches 0, evaluate the function $f\left(x\right)=\frac{\cos 2x-\cos x}{x^2}$ for the numbers $-0.5,-0.1,-0.01,-0.001,-0.0001,0.5,0.1,0.01,0.001\text{ and }0.0001$.

Evaluate the function (correct to 6 decimal places) for the values of x and get the following table of values.

$x$	$\cos 2x-\cos x$	$x^2$	$f\left(x\right)=\frac{\cos 2x-\cos x}{x^2}$
−0.5	−0.337280	0.25	−1.349121
−0.1	−0.014937	0.01	−1.493758
−0.01	−0.000149	0.0001	−1.499937
−0.001	−0.0000015	0.000001	−1.499999
−0.0001	−0.000000015	0.00000001	−1.499999
0.5	−0.337280	0.25	−1.349121
0.1	−0.014937	0.01	−1.493758
0.01	−0.000149	0.0001	−1.499937
0.001	−0.0000015	0.000001	−1.499999
0.0001	−0.000000015	0.00000001	−1.499999

Here, it is observed that the value of $f\left(x\right)$ gets closer to −1.5 as x approaches to 0 from the either side. That is, $\underset{x\to 0}{\lim }\frac{\cos 2x-\cos x}{x^2}=-1.5$.

Therefore, it is verified from the table that $\underset{x\to 0}{\lim }\frac{\cos 2x-\cos x}{x^2}=-1.5$.