Question

Which of the following functions have a graph which lies between the graphs of \(\mathrm{y}=|\mathrm{x}|\) and \(\mathrm{y}=-|\mathrm{x}|\) and have a limiting value as \(\mathrm{x} \rightarrow 0\). (A) \(\mathrm{y}=\mathrm{x} \cos \mathrm{x}\) (B) \(y=|x| \sin x\) (C) \(\mathrm{y}=\mathrm{x} \cos \frac{\mathrm{l}}{\mathrm{x}}\) (D) \(\mathrm{y}=\left|\mathrm{x} \sin \frac{1}{\mathrm{x}}\right|\)

Short answer

(A) \(y=x\cos x\) (B) \(y=|x|\sin x\) (C) \(y=x\cos \frac{1}{x}\) (D) \(y=|x\sin \frac{1}{x}|\) Answer: Functions (B) and (D) have graphs that lie between the graphs of \(y = |x|\) and \(y=-|x|\) and have a limiting value as \(x\) approaches 0.

Step-by-step solution

Examine function (A): \(y=x\cos x\)

First, we examine the values of the function as \(x\) approaches 0. For any value of \(x\), the value of \(\cos{x}\) lies in the range of \([-1, 1]\). As \(x \rightarrow 0\), \(y= x\cos{x}\) also approaches 0 because both \(x\) and \(\cos{x}\) are continuous near 0. The function \(x\cos{x}\) is continuous for all real numbers, so the output exists for any value of \(x\). However, as \(x\cos x\) can also take negative values in some intervals of \(x\), it doesn't lie between the graphs of \(y = |x|\) and \(y = -|x|\) for all \(x\). So the function (A) does not fulfill the requirements. #Step 2: Analyzing function B#

Examine function (B): \(y=|x|\sin x\)

For function (B), as \(x\) approaches 0, the function \(|x|\) also approaches 0. At the same time, the \(\sin x\) becomes 0 as \(x \rightarrow 0\). So, as \(x \rightarrow 0\), \(y=|x|\sin{x} \rightarrow 0\). Moreover, since the term \(|x|\) is non-negative, the values of the function \(|x|\sin x\) will always be non-negative, which means that it lies between the graphs of \(y = |x|\) and \(y = -|x|\). So the function (B) fulfills the requirements. #Step 3: Analyzing function C#

Examine function (C): \(y=x\cos \frac{1}{x}\)

For function (C), as \(x\) approaches 0, the term \(\cos \frac{1}{x}\) does not have a limit because it oscillates between -1 and 1 rapidly. Thus, there is no limiting value of this function when \(x \rightarrow 0\). So, the function (C) does not fulfill the requirements. #Step 4: Analyzing function D#

Examine function (D): \(y=|x\sin \frac{1}{x}|\)

For function (D), as \(x\) approaches 0, the term \(\sin \frac{1}{x}\) oscillates between -1 and 1, but it's multiplied by \(|x|\) which becomes 0 as \(x \rightarrow 0\). As a result, the function approaches 0 when \(x \rightarrow 0\) i.e., \(y=|x\sin \frac{1}{x}| \rightarrow 0\). Additionally, since the function involves the absolute value, it is non-negative; this means it lies between the graphs of \(y = |x|\) and \(y = -|x|\). So, the function (D) fulfills the requirements. #Conclusion# Functions (B) and (D) have graphs that lie between the graphs of \(y = |x|\) and \(y=-|x|\) and have a limiting value as \(x\) approaches 0.