If f(x) = begin{cases} frac{x ln(cos x)}{ln(1 | vedclass

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(C) To check continuity at $x = 0$,we evaluate $\lim_{x 	o 0} f(x)$.$\lim_{x 	o 0} \frac{x \ln(\cos x)}{\ln(1 + x^2)} = \lim_{x 	o 0} \frac{x \ln(1

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$f$ is differentiable at $x = 0$

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(C) To check continuity at $x = 0$,we evaluate $\lim_{x 	o 0} f(x)$.$\lim_{x 	o 0} \frac{x \ln(\cos x)}{\ln(1 + x^2)} = \lim_{x 	o 0} \frac{x \ln(1 + (\cos x - 1))}{\ln(1 + x^2)}$.Using the standard limit $\ln(1 + u) \approx u$ as $u 	o 0$:$\lim_{x 	o 0} \frac{x(\cos x - 1)}{x^2} = \lim_{x 	o 0} \frac{-\frac{x^2}{2}}{x} = \lim_{x 	o 0} -\frac{x}{2} = 0$.Since $\lim_{x 	o 0} f(x) = f(0) = 0$,$f$ is continuous at $x = 0$.To check differentiability,we evaluate $f'(0) = \lim_{h 	o 0} \frac{f(h) - f(0)}{h} = \lim_{h 	o 0} \frac{h \ln(\cos h)}{h \ln(1 + h^2)} = \lim_{h 	o 0} \frac{\ln(\cos h)}{\ln(1 + h^2)}$.Using $\ln(\cos h) \approx \ln(1 - \frac{h^2}{2}) \approx -\frac{h^2}{2}$ and $\ln(1 + h^2) \approx h^2$:$f'(0) = \lim_{h 	o 0} \frac{-\frac{h^2}{2}}{h^2} = -\frac{1}{2}$.Since the limit exists,$f$ is differentiable at $x = 0$.

If $f(x) = \begin{cases} \frac{x \ln(\cos x)}{\ln(1 + x^2)} & x \neq 0 \\ 0 & x = 0 \end{cases}$,then:

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