Given f(x) = begin{cases} frac{ln(1+text{sgn} | vedclass

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(C) To check continuity at $x = 0$,we evaluate the left-hand limit $(LHL)$ and right-hand limit $(RHL)$.For $x > 0$,$[x] = 0$,so $	ext{sgn}[x] = 0$.

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For $k = 2$,$f(x)$ has a non-removable discontinuity at $x = 0$

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(C) To check continuity at $x = 0$,we evaluate the left-hand limit $(LHL)$ and right-hand limit $(RHL)$.For $x > 0$,$[x] = 0$,so $	ext{sgn}[x] = 0$. Thus,$f(x) = \frac{\ln(1+x^2)}{1-\cos x}$.$RHL = \mathop {\lim }\limits_{h 	o 0^+} \frac{\ln(1+h^2)}{1-\cos h} = \mathop {\lim }\limits_{h 	o 0^+} \frac{\ln(1+h^2)}{h^2} \cdot \frac{h^2}{1-\cos h} = 1 \cdot 2 = 2$.For $x $LHL = \mathop {\lim }\limits_{h 	o 0^+} \frac{\ln(h^2)}{1-\cos(h)} = \mathop {\lim }\limits_{h 	o 0^+} \frac{\ln(h^2)}{h^2/2} = -\infty$.Since the $LHL$ does not exist (it is $-\infty$),the limit $\mathop {\lim }\limits_{x 	o 0} f(x)$ does not exist.Therefore,$f(x)$ has a non-removable discontinuity at $x = 0$ for any value of $k$.

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