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 \( x = 0 \)

For \( f(x) \) to be continuous at \( x = 0 \), we need to show that:

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

We know that \( f(0) = 0 \). Now we need to find \( \lim_{x \to 0} f(x) \):

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

As \( x \) approaches \( 0 \), \( \frac{1}{x} \) approaches infinity. The sine function oscillates between -1 and 1. Therefore, we can bound the limit:

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

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

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

Thus, we have:

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

This shows that \( f(x) \) is continuous at \( x = 0 \).

### Step 2: Check Differentiability at \( x = 0 \)

To check if \( f(x) \) is differentiable at \( x = 0 \), we need to find the derivative \( f'(0) \):

The derivative at \( x = 0 \) is defined as:

\[ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{f(h)}{h} \]

Substituting \( f(h) \):

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

As \( h \) approaches \( 0 \), \( \frac{1}{h} \) approaches infinity, and \( \sin\left(\frac{1}{h}\right) \) oscillates between -1 and 1. Therefore, the limit does not exist because it does not settle at a single value.

Thus, we conclude that \( f'(0) \) does not exist, meaning \( f(x) \) is not differentiable at \( x = 0 \).

### Final Conclusion

We have shown that \( f(x) \) is continuous at \( x = 0 \) and not differentiable at \( x = 0 \).

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