If f(x) = begin{cases} x^{alpha} sin left( fr | vedclass

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(C) Given,$f(x) = \begin{cases} x^{\alpha} \sin \left( \frac{1}{x} ight) ; & x 
eq 0 \ 0 ; & x = 0 \end{cases}$ For continuity at $x = 0$,$\lim_{x

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$f(x)$ is continuous and differentiable for $\alpha > 1$

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(C) Given,$f(x) = \begin{cases} x^{\alpha} \sin \left( \frac{1}{x} ight) ; & x 
eq 0 \ 0 ; & x = 0 \end{cases}$ For continuity at $x = 0$,$\lim_{x ightarrow 0} f(x) = f(0) = 0$. $\lim_{x ightarrow 0} x^{\alpha} \sin \left( \frac{1}{x} ight) = 0$ only if $\alpha > 0$. Thus,$f(x)$ is continuous for $\alpha > 0$. For differentiability at $x = 0$,$f^{\prime}(0) = \lim_{h ightarrow 0} \frac{f(0 + h) - f(0)}{h} = \lim_{h ightarrow 0} \frac{h^{\alpha} \sin \left( \frac{1}{h} ight) - 0}{h} = \lim_{h ightarrow 0} h^{\alpha - 1} \sin \left( \frac{1}{h} ight)$. This limit exists and is equal to $0$ if $\alpha - 1 > 0$,i.e.,$\alpha > 1$. Therefore,$f(x)$ is continuous and differentiable for $\alpha > 1$.

If $f(x) = \begin{cases} x^{\alpha} \sin \left( \frac{1}{x} \right), & x \neq 0 \\ 0, & x = 0 \end{cases}$; Which of the following is true?

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