Question

Let f : \(\R→\R\) be the function defined by
\(f(x)=\begin{cases} \lim\limits_{h \rightarrow0}\frac{(x+h)\sin(\frac{1}{x}+h)-x\sin\frac{1}{x}}{h} & x \ne 0\\ 0, & x=0\end{cases}\)
Then which one of the following statements is NOT true ?

• A. \(f(\frac{2}{\pi})=1\)
• B. \(f(\frac{1}{\pi})=\frac{1}{\pi}\)
• C. \(f(-\frac{2}{\pi})=-1\)
• D. f is not continuous at x = 0

Answer 1

The Correct Option is B

The problem involves evaluating the behavior of a function \( f:\mathbb{R} \to \mathbb{R} \) defined in a piecewise manner, with a crucial point of interest being the limit involved in its definition. We need to analyze each statement and determine which one is not true.

1. Analyze the definition of the function \( f(x) \):

   The function \( f(x) \) is defined as: \( f(x) = \begin{cases} \lim\limits_{h \rightarrow 0} \frac{(x+h)\sin(\frac{1}{x}+h)-x\sin\frac{1}{x}}{h}, & x \ne 0 \\ 0, & x = 0 \end{cases} \)

2. Compute the derivative for \( x \neq 0 \), using the standard derivative rules and limits:

   For \( x \neq 0 \), the expression is essentially the definition of a derivative of \( f(x) = x \sin(\frac{1}{x}) \). The derivative would be: \[ \frac{d}{dx}(x \sin(\frac{1}{x})) = \sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right) \cdot \frac{1}{x} \]

3. Evaluate the statements:
   • For \( f\left(\frac{2}{\pi}\right) = 1 \):

     Substitute \( x = \frac{2}{\pi} \) in the derivative expression: \[ f\left(\frac{2}{\pi}\right) = \sin(\pi) - \cos(\pi) \cdot \frac{\pi}{2} = 0 - (-1) \cdot \frac{\pi}{2} = 1 \] Therefore, this statement is true.

   • For \( f\left(\frac{1}{\pi}\right) = \frac{1}{\pi} \):

     Substitute \( x = \frac{1}{\pi} \) in the derivative expression: \[ f\left(\frac{1}{\pi}\right) = \sin(1) - \cos(1) \cdot \pi \] Since \(\sin(1) - \cos(1) \cdot \pi \neq \frac{1}{\pi}\), this statement is not true.

   • For \( f\left(-\frac{2}{\pi}\right) = -1 \):

     Substitute \( x = -\frac{2}{\pi} \) in the derivative expression: \[ f\left(-\frac{2}{\pi}\right) = \sin(-\pi) - \cos(-\pi) \cdot \frac{\pi}{2} = 0 - 1 \cdot \frac{\pi}{2} = -1 \] Therefore, this statement is true.

   • For continuity at \( x = 0 \):

     The function value at \( x = 0 \) is \( f(0) = 0 \). To check continuity, evaluate the limit: \[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \left(\sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right) \cdot \frac{1}{x}\right) \] This does not exist, indicating discontinuity at \( x = 0 \). Therefore, this statement is true.

Conclusion: The statement \(f(\frac{1}{\pi})=\frac{1}{\pi}\) is NOT true.