The function f defined by:
`f(x) =x{1+ 1/3 sin (log x^(2))}, x ne 0` and `f(0) =0`, is everywhere continuous but has no differential coefficient at the origin.

Answer

To determine whether the function \( f(x) = x\left(1 + \frac{1}{3} \sin(\log x^2)\right) \) for \( x \neq 0 \) and \( f(0) = 0 \) is continuous everywhere and has no differential coefficient at the origin, we will follow these steps:

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

To check if \( f \) is continuous at \( x = 0 \), we need to evaluate:

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

We know that:

\[ f(x) = x\left(1 + \frac{1}{3} \sin(\log x^2)\right) \]

As \( x \to 0 \), \( \log x^2 \to -\infty \). The sine function oscillates between -1 and 1, so:

\[ \sin(\log x^2) \text{ oscillates between } -1 \text{ and } 1. \]

Thus, we can write:

\[ -1 \leq \sin(\log x^2) \leq 1 \implies -\frac{1}{3} \leq \frac{1}{3} \sin(\log x^2) \leq \frac{1}{3}. \]

Adding 1 to all parts gives:

\[ \frac{2}{3} \leq 1 + \frac{1}{3} \sin(\log x^2) \leq \frac{4}{3}. \]

Now, we can multiply by \( x \):

\[ \frac{2}{3}x \leq f(x) \leq \frac{4}{3}x. \]

Taking the limit as \( x \to 0 \):

\[ \lim_{x \to 0} \frac{2}{3}x = 0 \quad \text{and} \quad \lim_{x \to 0} \frac{4}{3}x = 0. \]

By the Squeeze Theorem:

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

Since \( f(0) = 0 \), we conclude that \( f \) is continuous at \( x = 0 \).

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

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

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

Substituting \( f(h) \):

\[ f(h) = h\left(1 + \frac{1}{3} \sin(\log h^2)\right). \]

Thus,

\[ \frac{f(h)}{h} = 1 + \frac{1}{3} \sin(\log h^2). \]

As \( h \to 0 \), \( \log h^2 \to -\infty \), causing \( \sin(\log h^2) \) to oscillate between -1 and 1. Therefore, the limit does not exist as it oscillates:

\[ 1 + \frac{1}{3} \sin(\log h^2) \text{ oscillates between } \frac{2}{3} \text{ and } \frac{4}{3}. \]

Since the limit does not approach a single value, \( f'(0) \) does not exist.

### Conclusion

The function \( f \) is continuous everywhere but does not have a differential coefficient at the origin.

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