Question

(a) By graphing the function \(f\left( x \right)={\left( cos2x-cosx \right)}/{{{x}^{2}}}\;\)and zooming in toward the point where the graph crosses the y-axis, estimate the value of \(\mathop {lim}\limits_{x \to 0} f\left( x \right)\). (b) Check your answer in part (a) by evaluating f(x) for values of x that approach 0.

Short answer

(a) \(\underset{x\to 0}{\mathop{\lim }}\,=\frac{\left( cos2x-cosx \right)}{{{x}^{2}}}=-1.5\)

(b) By the table the answer is the same so the graph is right.

Step-by-step solution

Introduction

In one sided limit,

\(\mathop {\lim }\limits_{x \to {0^ - }} H(t) = 0{\rm{ }}and{\rm{ }}\mathop {\lim }\limits_{x \to {0^ + }} H(t) = 1\)

The symbol \(x \to {0^ - }{\rm{ and }}x \to {0^ + }{\rm{ }}\) indicates that considered values are less than zero and greater than zero respectively.

The negative sign (-) value will be get from the left side and the positive sign value (+)will be get from the right side.

If both the values will be equal then than the value of \(x \to 0\)will be also equal to the both the values and if both the values are unequal then \(x \to 0\)won’t be equal to any of them.

Given information

The given function is, \(f\left( x \right)={\left( \cos 2x-\cos x \right)}/{{{x}^{2}}}\;\).

(a) Graph

Now, draw the graph of the function \(f\left( x \right)={\left( \cos 2x-\cos x \right)}/{{{x}^{2}}}\;\) by using CAS.

From the graph it is clear that \(\underset{x\to 0}{\mathop{\lim }}\,=\frac{\left( cos2x-cosx \right)}{{{x}^{2}}}=-1.5\).

(b) Answer checking

By putting the different values of x in the right and left limit of function \(f\left( x \right)={\left( cos2x-cosx \right)}/{{{x}^{2}}}\;\).

x	f(x)
\(\begin{aligned}{l} \pm 0.1\\ \pm 0.01\\ \pm 0.001\\ \pm 0.001\end{aligned}\)	\(\begin{aligned}{l} - 1.493759\\ - 1.499938\\ - 1.499999\\ - 1.500000\end{aligned}\)

By the table the answer of the graph is right that is \( - 1.5\).