Key Concepts and Formulas

• Continuity at a point $x=a$: A function $f(x)$ is continuous at $x=a$ if $\lim_{x \to a} f(x) = f(a)$.
• Differentiability at a point $x=a$: A function $f(x)$ is differentiable at $x=a$ if the limit of the difference quotient exists: $f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$.
• Continuity of the derivative $f'(x)$ at $x=a$: The derivative $f'(x)$ is continuous at $x=a$ if $\lim_{x \to a} f'(x) = f'(a)$.
• Limit of $x^n \sin(1/x)$ and $x^n \cos(1/x)$ as $x \to 0$: For any $n > 0$, $\lim_{x \to 0} x^n \sin(1/x) = 0$ and $\lim_{x \to 0} x^n \cos(1/x) = 0$. This is because $|\sin(1/x)| \le 1$ and $|\cos(1/x)| \le 1$.

Step-by-Step Solution

Step 1: Check the continuity of $f(x)$ at $x=0$. To check for continuity at $x=0$, we need to verify if $\lim_{x \to 0} f(x) = f(0)$. Given $f(0) = 0$. Now, let's calculate the limit: $\lim_{x \to 0} f(x) = \lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right)$ We know that $-1 \le \sin\left(\frac{1}{x}\right) \le 1$ for all $x \ne 0$. Multiplying by $x^2$ (which is non-negative): $-x^2 \le x^2 \sin\left(\frac{1}{x}\right) \le x^2$ As $x \to 0$, we have $\lim_{x \to 0} (-x^2) = 0$ and $\lim_{x \to 0} x^2 = 0$. By the Squeeze Theorem, $\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0$ Since $\lim_{x \to 0} f(x) = 0 = f(0)$, the function $f(x)$ is continuous at $x=0$.

Step 2: Check the differentiability of $f(x)$ at $x=0$. To check for differentiability at $x=0$, we need to evaluate the limit of the difference quotient: $f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h}$ Substituting the function definition: $f'(0) = \lim_{h \to 0} \frac{f(h) - 0}{h} = \lim_{h \to 0} \frac{h^2 \sin\left(\frac{1}{h}\right)}{h}$ $f'(0) = \lim_{h \to 0} h \sin\left(\frac{1}{h}\right)$ Using the Squeeze Theorem again, since $-1 \le \sin\left(\frac{1}{h}\right) \le 1$, we have: $-|h| \le h \sin\left(\frac{1}{h}\right) \le |h|$ As $h \to 0$, $\lim_{h \to 0} (-|h|) = 0$ and $\lim_{h \to 0} |h| = 0$. Therefore, by the Squeeze Theorem, $f'(0) = 0$ Since the limit exists, $f(x)$ is differentiable at $x=0$, and its derivative at $x=0$ is $0$.

Step 3: Find the derivative $f'(x)$ for $x \ne 0$. For $x \ne 0$, $f(x) = x^2 \sin\left(\frac{1}{x}\right)$. We use the product rule and the chain rule to find $f'(x)$. Let $u = x^2$ and $v = \sin\left(\frac{1}{x}\right)$. Then $\frac{du}{dx} = 2x$. To find $\frac{dv}{dx}$, let $w = \frac{1}{x} = x^{-1}$. Then $\frac{dw}{dx} = -x^{-2} = -\frac{1}{x^2}$. So, $\frac{dv}{dx} = \cos(w) \cdot \frac{dw}{dx} = \cos\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right) = -\frac{1}{x^2}\cos\left(\frac{1}{x}\right)$. Using the product rule, $f'(x) = \frac{du}{dx}v + u\frac{dv}{dx}$: $f'(x) = (2x) \sin\left(\frac{1}{x}\right) + x^2 \left(-\frac{1}{x^2}\cos\left(\frac{1}{x}\right)\right)$ $f'(x) = 2x \sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right) \quad \text{for } x \ne 0$

Step 4: Check the continuity of $f'(x)$ at $x=0$. For $f'(x)$ to be continuous at $x=0$, we need $\lim_{x \to 0} f'(x) = f'(0)$. We found $f'(0) = 0$ in Step 2. Now, let's evaluate the limit of $f'(x)$ as $x \to 0$: $\lim_{x \to 0} f'(x) = \lim_{x \to 0} \left( 2x \sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right) \right)$ We can split this into two limits: $\lim_{x \to 0} 2x \sin\left(\frac{1}{x}\right) - \lim_{x \to 0} \cos\left(\frac{1}{x}\right)$ For the first limit, using the Squeeze Theorem as in Step 1: $\lim_{x \to 0} 2x \sin\left(\frac{1}{x}\right) = 0$ For the second limit, $\lim_{x \to 0} \cos\left(\frac{1}{x}\right)$: As $x$ approaches $0$, $\frac{1}{x}$ approaches $\pm \infty$. The cosine function oscillates infinitely between $-1$ and $1$ as its argument goes to infinity. Therefore, the limit $\lim_{x \to 0} \cos\left(\frac{1}{x}\right)$ does not exist. Since one part of the limit does not exist, the entire limit $\lim_{x \to 0} f'(x)$ does not exist.

Step 5: Conclude the continuity of $f(x)$ and $f'(x)$ at $x=0$. From Step 1, we concluded that $f(x)$ is continuous at $x=0$. From Step 2, we concluded that $f(x)$ is differentiable at $x=0$, and $f'(0)=0$. From Step 4, we concluded that $\lim_{x \to 0} f'(x)$ does not exist. For $f'(x)$ to be continuous at $x=0$, we need $\lim_{x \to 0} f'(x) = f'(0)$. Since the limit does not exist, $f'(x)$ is not continuous at $x=0$.

Common Mistakes & Tips

• Confusing differentiability with the existence of the derivative formula: While the formula $f'(x) = 2x \sin(1/x) - \cos(1/x)$ is valid for $x \ne 0$, applying it directly to find the limit as $x \to 0$ without checking the differentiability at $x=0$ can lead to errors. Always check differentiability at the specific point using the definition of the derivative.
• Assuming continuity of $f'(x)$ from differentiability of $f(x)$: A function can be differentiable at a point, but its derivative may not be continuous at that point, as demonstrated in this problem. The existence of $f'(a)$ does not guarantee the continuity of $f'(x)$ at $x=a$.
• Misinterpreting limits involving $\sin(1/x)$ or $\cos(1/x)$: The limits of $\sin(1/x)$ and $\cos(1/x)$ as $x \to 0$ do not exist. However, when multiplied by a term that goes to zero faster than $1/x$ (like $x$ or $x^2$), the limit can exist (and is often 0), as shown by the Squeeze Theorem.

Summary

We first established that the function $f(x)$ is continuous at $x=0$ by verifying that $\lim_{x \to 0} f(x) = f(0)$. Next, we found that $f(x)$ is differentiable at $x=0$ using the definition of the derivative, yielding $f'(0) = 0$. We then derived the expression for $f'(x)$ for $x \ne 0$. Upon examining the limit of $f'(x)$ as $x \to 0$, we found that $\lim_{x \to 0} f'(x)$ does not exist due to the $\cos(1/x)$ term. Since $\lim_{x \to 0} f'(x)$ does not exist, $f'(x)$ is not continuous at $x=0$. Therefore, $f(x)$ is continuous at $x=0$, but its derivative $f'(x)$ is not continuous at $x=0$.

The final answer is \boxed{A}.