Show that `f(x)={{:("x sin"(1)/(x)",","when",x ne 0),(0",","when",x = 0):}` is continuous but not differentiable at x = 0

Answer

To show that the function \( f(x) = \begin{cases} x \sin\left(\frac{1}{x}\right) & \text{when } x \neq 0 \\ 0 & \text{when } x = 0 \end{cases} \) is continuous but not differentiable at \( x = 0 \), we will follow these steps:

### Step 1: Check Continuity at \( x = 0 \)

To prove that \( f(x) \) is continuous at \( x = 0 \), we need to show that:

\[ \lim_{x \to 0} f(x) = f(0) \]

Since \( f(0) = 0 \), we need to evaluate \( \lim_{x \to 0} f(x) \):

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} x \sin\left(\frac{1}{x}\right) \]

### Step 2: Evaluate the Limit

We know that \( \sin\left(\frac{1}{x}\right) \) oscillates between -1 and 1 for all \( x \neq 0 \). Therefore, we can bound \( f(x) \):

\[ -x \leq x \sin\left(\frac{1}{x}\right) \leq x \]

As \( x \) approaches 0, both \( -x \) and \( x \) approach 0. By the Squeeze Theorem:

\[ \lim_{x \to 0} x \sin\left(\frac{1}{x}\right) = 0 \]

### Step 3: Conclude Continuity

Thus, we have:

\[ \lim_{x \to 0} f(x) = 0 = f(0) \]

Hence, \( f(x) \) is continuous at \( x = 0 \).

### Step 4: Check Differentiability at \( x = 0 \)

To check if \( f(x) \) is differentiable at \( x = 0 \), we need to find the left-hand derivative and the right-hand derivative.

#### Left-Hand Derivative

The left-hand derivative at \( x = 0 \) is given by:

\[ f'_{-}(0) = \lim_{h \to 0^-} \frac{f(0) - f(-h)}{-h} = \lim_{h \to 0^-} \frac{0 - (-h \sin\left(-\frac{1}{h}\right))}{-h} \]

This simplifies to:

\[ = \lim_{h \to 0^-} \sin\left(-\frac{1}{h}\right) = -\lim_{h \to 0^-} \sin\left(\frac{1}{h}\right) \]

Since \( \sin\left(\frac{1}{h}\right) \) oscillates between -1 and 1, the limit does not exist.

#### Right-Hand Derivative

The right-hand derivative at \( x = 0 \) is given by:

\[ f'_{+}(0) = \lim_{h \to 0^+} \frac{f(0) - f(h)}{-h} = \lim_{h \to 0^+} \frac{0 - (h \sin\left(\frac{1}{h}\right))}{-h} \]

This simplifies to:

\[ = \lim_{h \to 0^+} -\sin\left(\frac{1}{h}\right) \]

Again, since \( \sin\left(\frac{1}{h}\right) \) oscillates between -1 and 1, this limit does not exist.

### Step 5: Conclude Non-Differentiability

Since the left-hand and right-hand derivatives do not exist, \( f(x) \) is not differentiable at \( x = 0 \).

### Final Conclusion

Thus, we have shown that \( f(x) \) is continuous at \( x = 0 \) but not differentiable at \( x = 0 \).

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