Question

For \( x \in \mathbb{R} \), let \( f(x) = \begin{cases} x^3 \sin \left( \frac{1}{x} \right), & x \neq 0 \\ 0, & x = 0 \end{cases} \) . Then which one of the following is FALSE?
 

• A. \( \lim_{x \to 0} \dfrac{f(x)}{x} = 0 \)
• B. \( \lim_{x \to 0} \dfrac{f(x)}{x^2} = 0 \)
• C. \( \dfrac{f(x)}{x^2} \) has infinitely many maxima and minima on the interval \( (0,1) \)
• D. \( \dfrac{f(x)}{x^4} \) is continuous at \( x = 0 \) but not differentiable at \( x = 0 \)

Hint:

For functions involving oscillatory terms like \( \sin \left( \frac{1}{x} \right) \), check for limits, continuity, and differentiability carefully by analyzing the behavior near critical points like \( x = 0 \).

Answer 1

The Correct Option is D

Given function: $$f(x) = \begin{cases} x^3\sin\left(\frac{1}{x}\right), & x \neq 0 \ 0, & x = 0 \end{cases}$$

Option (A): $\lim_{x \to 0} \frac{f(x)}{x} = 0$

For $x \neq 0$: $$\frac{f(x)}{x} = \frac{x^3\sin\left(\frac{1}{x}\right)}{x} = x^2\sin\left(\frac{1}{x}\right)$$

Since $\left|\sin\left(\frac{1}{x}\right)\right| \leq 1$: $$\left|x^2\sin\left(\frac{1}{x}\right)\right| \leq |x^2| \to 0 \text{ as } x \to 0$$

By the squeeze theorem: $$\lim_{x \to 0} \frac{f(x)}{x} = 0$$

Option (A) is TRUE 

Option (B): $\lim_{x \to 0} \frac{f(x)}{x^2} = 0$

For $x \neq 0$: $$\frac{f(x)}{x^2} = \frac{x^3\sin\left(\frac{1}{x}\right)}{x^2} = x\sin\left(\frac{1}{x}\right)$$

Since $\left|\sin\left(\frac{1}{x}\right)\right| \leq 1$: $$\left|x\sin\left(\frac{1}{x}\right)\right| \leq |x| \to 0 \text{ as } x \to 0$$

By the squeeze theorem: $$\lim_{x \to 0} \frac{f(x)}{x^2} = 0$$

Option (B) is TRUE 

Option (C): $\frac{f(x)}{x^2}$ has infinitely many maxima and minima on $(0,1)$

For $x \in (0,1)$: $$\frac{f(x)}{x^2} = x\sin\left(\frac{1}{x}\right)$$

Taking the derivative: $$\frac{d}{dx}\left[x\sin\left(\frac{1}{x}\right)\right] = \sin\left(\frac{1}{x}\right) + x\cos\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right)$$ $$= \sin\left(\frac{1}{x}\right) - \frac{1}{x}\cos\left(\frac{1}{x}\right)$$

Critical points occur when: $$\sin\left(\frac{1}{x}\right) = \frac{1}{x}\cos\left(\frac{1}{x}\right)$$ $$\tan\left(\frac{1}{x}\right) = \frac{1}{x}$$

As $x \to 0^+$, $\frac{1}{x} \to +\infty$, and $\sin\left(\frac{1}{x}\right)$ oscillates infinitely many times. This creates infinitely many critical points in $(0,1)$, leading to infinitely many local maxima and minima.

Option (C) is TRUE 

Option (D): $\frac{f(x)}{x^4}$ is continuous at $x = 0$ but not differentiable at $x = 0$

For $x \neq 0$: $$\frac{f(x)}{x^4} = \frac{x^3\sin\left(\frac{1}{x}\right)}{x^4} = \frac{\sin\left(\frac{1}{x}\right)}{x}$$

Check continuity at $x = 0$:

We need to check if $\lim_{x \to 0} \frac{f(x)}{x^4} = \frac{f(0)}{0^4}$.

Since $f(0) = 0$, we need $\lim_{x \to 0} \frac{f(x)}{x^4} = 0$ for continuity.

$$\lim_{x \to 0} \frac{\sin\left(\frac{1}{x}\right)}{x}$$

As $x \to 0$, $\frac{1}{x} \to \infty$ and $\sin\left(\frac{1}{x}\right)$ oscillates between $-1$ and $1$.

Therefore: $\frac{\sin\left(\frac{1}{x}\right)}{x}$ oscillates and grows without bound as $x \to 0$.

The limit does not exist, so $\frac{f(x)}{x^4}$ is NOT continuous at $x = 0$.

Option (D) is FALSE 

Answer: (D)