Key Concepts and Formulas

• Definition of Differentiability: A function $f(x)$ is differentiable at a point $x=c$ if the limit of the difference quotient exists: $f'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h}$ This requires the left-hand derivative (LHD) and the right-hand derivative (RHD) to be equal: $LHD = \lim_{h \to 0^-} \frac{f(c+h) - f(c)}{h} = \lim_{h \to 0^+} \frac{f(c+h) - f(c)}{h} = RHD$
• Limit of $\sin(1/x)$ as $x \to 0$: The limit of $\sin(1/x)$ as $x \to 0$ does not exist because the function oscillates infinitely often between -1 and 1 as $x$ approaches 0. However, the limit of $x \sin(1/x)$ as $x \to 0$ is 0, due to the Squeeze Theorem, since $-|x| \le x \sin(1/x) \le |x|$.
• Derivative of a product: If $f(x) = g(x)h(x)$, then $f'(x) = g'(x)h(x) + g(x)h'(x)$.

Step-by-Step Solution

Step 1: Analyze differentiability at $x=1$. To check for differentiability at $x=1$, we need to evaluate the right-hand derivative (RHD) and the left-hand derivative (LHD) at $x=1$.

• Calculate the Right-Hand Derivative (RHD) at $x=1$: The RHD is given by: $Rf'(1) = \lim_{h \to 0^+} \frac{f(1+h) - f(1)}{h}$ Substitute the definition of $f(x)$: For $x \ne 1$, $f(x) = (x-1)\sin\left(\frac{1}{x-1}\right)$, and $f(1) = 0$. $Rf'(1) = \lim_{h \to 0^+} \frac{((1+h)-1)\sin\left(\frac{1}{(1+h)-1}\right) - 0}{h}$ $Rf'(1) = \lim_{h \to 0^+} \frac{h\sin\left(\frac{1}{h}\right)}{h}$ $Rf'(1) = \lim_{h \to 0^+} \sin\left(\frac{1}{h}\right)$ As $h \to 0^+$, $\frac{1}{h} \to \infty$. The function $\sin(y)$ oscillates between -1 and 1 as $y \to \infty$. Therefore, the limit $\lim_{h \to 0^+} \sin\left(\frac{1}{h}\right)$ does not exist.

• Calculate the Left-Hand Derivative (LHD) at $x=1$: The LHD is given by: $Lf'(1) = \lim_{h \to 0^-} \frac{f(1+h) - f(1)}{h}$ Substitute the definition of $f(x)$: For $x \ne 1$, $f(x) = (x-1)\sin\left(\frac{1}{x-1}\right)$, and $f(1) = 0$. $Lf'(1) = \lim_{h \to 0^-} \frac{((1+h)-1)\sin\left(\frac{1}{(1+h)-1}\right) - 0}{h}$ $Lf'(1) = \lim_{h \to 0^-} \frac{h\sin\left(\frac{1}{h}\right)}{h}$ $Lf'(1) = \lim_{h \to 0^-} \sin\left(\frac{1}{h}\right)$ As $h \to 0^-$, $\frac{1}{h} \to -\infty$. The function $\sin(y)$ oscillates between -1 and 1 as $y \to -\infty$. Therefore, the limit $\lim_{h \to 0^-} \sin\left(\frac{1}{h}\right)$ does not exist.

Since both the RHD and LHD at $x=1$ do not exist, the function $f(x)$ is not differentiable at $x=1$.

Step 2: Analyze differentiability at $x=0$. To check for differentiability at $x=0$, we need to evaluate the limit of the difference quotient at $x=0$. Since $x=0$ is not the point where the function definition changes ($x=1$), we can use the general form of the derivative for $x \ne 1$.

• Calculate the derivative for $x \ne 1$: For $x \ne 1$, $f(x) = (x-1)\sin\left(\frac{1}{x-1}\right)$. We use the product rule: $f'(x) = \frac{d}{dx}(x-1) \cdot \sin\left(\frac{1}{x-1}\right) + (x-1) \cdot \frac{d}{dx}\left(\sin\left(\frac{1}{x-1}\right)\right)$ $f'(x) = 1 \cdot \sin\left(\frac{1}{x-1}\right) + (x-1) \cdot \cos\left(\frac{1}{x-1}\right) \cdot \frac{d}{dx}\left(\frac{1}{x-1}\right)$ $f'(x) = \sin\left(\frac{1}{x-1}\right) + (x-1) \cdot \cos\left(\frac{1}{x-1}\right) \cdot \left(-\frac{1}{(x-1)^2}\right)$ $f'(x) = \sin\left(\frac{1}{x-1}\right) - \frac{1}{x-1}\cos\left(\frac{1}{x-1}\right)$

• Evaluate the derivative at $x=0$: We need to find $f'(0)$ using the definition of the derivative, or by taking the limit of $f'(x)$ as $x \to 0$ if the derivative is continuous at $x=0$. Let's use the definition of the derivative at $x=0$: $f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h}$ Since $0 \ne 1$, $f(0) = (0-1)\sin\left(\frac{1}{0-1}\right) = (-1)\sin(-1) = \sin(1)$. $f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h}$ For $h \ne 1$ (which is true as $h \to 0$), $f(h) = (h-1)\sin\left(\frac{1}{h-1}\right)$. $f'(0) = \lim_{h \to 0} \frac{(h-1)\sin\left(\frac{1}{h-1}\right) - \sin(1)}{h}$ This limit is hard to evaluate directly. Let's try to evaluate the limit of $f'(x)$ as $x \to 0$. $f'(x) = \sin\left(\frac{1}{x-1}\right) - \frac{1}{x-1}\cos\left(\frac{1}{x-1}\right)$ As $x \to 0$: $\sin\left(\frac{1}{x-1}\right) \to \sin\left(\frac{1}{0-1}\right) = \sin(-1) = -\sin(1)$. $\frac{1}{x-1} \to \frac{1}{0-1} = -1$. $\cos\left(\frac{1}{x-1}\right) \to \cos\left(\frac{1}{0-1}\right) = \cos(-1) = \cos(1)$. So, $\lim_{x \to 0} f'(x) = -\sin(1) - (-1)\cos(1) = -\sin(1) + \cos(1)$.

  Now we need to confirm if $f'(0)$ is indeed equal to this limit. We can use the definition of the derivative at $x=0$: $f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h}$ For $h \ne 1$, $f(h) = (h-1)\sin\left(\frac{1}{h-1}\right)$. And $f(0) = (0-1)\sin\left(\frac{1}{0-1}\right) = (-1)\sin(-1) = \sin(1)$. $f'(0) = \lim_{h \to 0} \frac{(h-1)\sin\left(\frac{1}{h-1}\right) - \sin(1)}{h}$ Let's try to evaluate the limit of the derivative expression we found: $f'(x) = \sin\left(\frac{1}{x-1}\right) - \frac{1}{x-1}\cos\left(\frac{1}{x-1}\right)$ As $x \to 0$, this expression approaches $\sin(-1) - (-1)\cos(-1) = -\sin(1) + \cos(1)$. Since the derivative $f'(x)$ exists for $x \ne 1$ and $\lim_{x \to 0} f'(x)$ exists and is finite, and also $f(x)$ is continuous at $x=0$, it implies that $f(x)$ is differentiable at $x=0$ and $f'(0) = \lim_{x \to 0} f'(x) = \cos(1) - \sin(1)$.

  Therefore, $f$ is differentiable at $x=0$.

Step 3: Conclude the differentiability. From Step 1, we found that $f$ is not differentiable at $x=1$. From Step 2, we found that $f$ is differentiable at $x=0$.

Combining these results, $f$ is differentiable at $x=0$ but not at $x=1$.

Common Mistakes & Tips

• Confusing the limit of $\sin(1/x)$ with $\sin(x)$: The limit of $\sin(1/x)$ as $x \to 0$ does not exist, but the limit of $x \sin(1/x)$ as $x \to 0$ is 0. This distinction is crucial when evaluating derivatives involving such terms.
• Assuming continuity implies differentiability: A function must be continuous at a point to be differentiable at that point. However, continuity does not guarantee differentiability (e.g., $|x|$ at $x=0$). Always check the definition of the derivative.
• Incorrect application of derivative rules at boundary points: When checking differentiability at a point where the function definition changes (like $x=1$ in this case), always use the limit definition of the derivative (RHD and LHD). Do not directly substitute into the derived formula for $f'(x)$ if that formula is only valid for $x \ne 1$.

Summary

We analyzed the differentiability of the function $f(x)$ at $x=1$ by calculating the right-hand and left-hand derivatives. We found that both limits do not exist, concluding that $f(x)$ is not differentiable at $x=1$. Next, we analyzed the differentiability at $x=0$. We first found the expression for the derivative $f'(x)$ for $x \ne 1$ using the product rule. Then, we evaluated the limit of $f'(x)$ as $x \to 0$, which yielded a finite value. This, combined with the continuity of $f(x)$ at $x=0$, implies that $f(x)$ is differentiable at $x=0$ and $f'(0) = \cos(1) - \sin(1)$. Therefore, the function is differentiable at $x=0$ but not at $x=1$.

The final answer is \boxed{A}.