Left-hand and Right-hand Derivatives

For continuity at $$x = 0$$:

Left limit: $$\lim_{x \to 0^-} \frac{\sin(ax)}{x} = a$$

Right limit: $$\lim_{x \to 0^+} \frac{e^{bx} - 1}{x} = b$$

For continuity: $$a = b = f(0)$$

For differentiability at $$x = 0$$:

Left derivative: $$f'_-(0) = \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{\sin(ah)/h - a}{h}$$

$$= \lim_{h \to 0^-} \frac{\sin(ah) - ah}{h^2}$$

Using Taylor expansion $$\sin(ah) = ah - \frac{a^3h^3}{6} + O(h^5)$$:

$$= \lim_{h \to 0^-} \frac{-\frac{a^3h^3}{6}}{h^2} = \lim_{h \to 0^-} \frac{-a^3h}{6} = 0$$

Right derivative: $$f'_+(0) = \lim_{h \to 0^+} \frac{e^{bh} - 1)/h - b}{h}$$

$$= \lim_{h \to 0^+} \frac{e^{bh} - 1 - bh}{h^2}$$

Using Taylor expansion $$e^{bh} = 1 + bh + \frac{b^2h^2}{2} + \frac{b^3h^3}{6} + ...$$:

$$= \lim_{h \to 0^+} \frac{\frac{b^2h^2}{2} + \frac{b^3h^3}{6} + ...}{h^2} = \frac{b^2}{2}$$

This directly requires:
$$b^2 = 0 \Rightarrow b = 0$$

Since continuity established that $$a = b$$, we find:
$$a = 0$$

\noindent Therefore, the values required for differentiability are $$a = 0$$ and $$b = 0$$, which yields a constant derivative value of $$f'(0) = 0$$.

We evaluate the left and right behaviors around the origin.
For $$x > 0$$, $$\text{sgn}(x) = 1 \implies f(x) = \frac{\ln(1 + \sin^2 x)}{x}$$.
Using Taylor expansions as $$x \to 0^+$$: $$\ln(1 + \sin^2 x) \approx \sin^2 x \approx x^2$$.
$$RHL = \lim_{x \to 0^+} \frac{x^2}{x} = 0$$.

For $$x < 0$$, $$\text{sgn}(x) = -1 \implies f(x) = \frac{\ln(1 - \sin^2 x)}{x} = \frac{\ln(\cos^2 x)}{x} = \frac{2 \ln(\cos x)}{x}$$.
Using expansions near zero: $$\ln(\cos x) \approx \ln(1 - \frac{x^2}{2}) \approx -\frac{x^2}{2}$$.
$$LHL = \lim_{x \to 0^-} \frac{2(-\frac{x^2}{2})}{x} = \lim_{x \to 0^-} (-x) = 0$$.
Since $$LHL = RHL = f(0) = 0$$, the function is continuous at $$x = 0$$.

Now evaluate differentiability using first principles:
$$RHD = \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{\ln(1 + \sin^2 h)}{h^2} = 1$$.
$$LHD = \lim_{h \to 0^+} \frac{f(-h) - f(0)}{-h} = \lim_{h \to 0^+} \frac{\frac{2\ln(\cos(-h))}{-h} - 0}{-h} = \lim_{h \to 0^+} \frac{2\ln(\cos h)}{h^2}$$.
Using our expansion: $$\lim_{h \to 0^+} \frac{2(-\frac{h^2}{2})}{h^2} = -1$$.
Since $$RHD = 1$$ and $$LHD = -1$$, the derivatives do not match. The function is continuous but not differentiable.