If f(x) = begin{cases} frac{x^2 ln cos x}{ln | vedclass

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(C) First,we check for continuity at $x=0$: $\lim_{x 	o 0} f(x) = \lim_{x 	o 0} \frac{x^2 \ln \cos x}{\ln (1+x^2)}$.Using standard limits $\lim_{u \

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differentiable at zero

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(C) First,we check for continuity at $x=0$: $\lim_{x 	o 0} f(x) = \lim_{x 	o 0} \frac{x^2 \ln \cos x}{\ln (1+x^2)}$.Using standard limits $\lim_{u 	o 0} \frac{\ln(1+u)}{u} = 1$,we have $\lim_{x 	o 0} \frac{\ln(1+x^2)}{x^2} = 1$.Thus,$\lim_{x 	o 0} f(x) = \lim_{x 	o 0} \left( \frac{x^2}{\ln(1+x^2)} ight) \cdot \ln \cos x = 1 \cdot \ln(1) = 0$.Since $\lim_{x 	o 0} f(x) = f(0) = 0$,the function is continuous at $x=0$.Now,we check for differentiability at $x=0$ using the definition $f'(0) = \lim_{h 	o 0} \frac{f(h) - f(0)}{h} = \lim_{h 	o 0} \frac{h^2 \ln \cos h}{h \ln(1+h^2)} = \lim_{h 	o 0} \frac{h \ln \cos h}{\ln(1+h^2)}$.Dividing numerator and denominator by $h^2$: $\lim_{h 	o 0} \frac{\frac{\ln \cos h}{h}}{\frac{\ln(1+h^2)}{h^2}} = \lim_{h 	o 0} \frac{\frac{\ln \cos h}{h}}{1}$.Using $L$'Hospital's rule on $\frac{\ln \cos h}{h}$: $\lim_{h 	o 0} \frac{-	an h}{1} = 0$.Since the limit exists and is equal to $0$,$f(x)$ is differentiable at $x=0$.

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

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