The value of k which makes f(x) = begin{cases | vedclass

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(D) For a function $f(x)$ to be continuous at $x = 0$,the limit $\lim_{x 	o 0} f(x)$ must exist and be equal to $f(0)$.Here,$f(0) = k$.We examine the

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None of these

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(D) For a function $f(x)$ to be continuous at $x = 0$,the limit $\lim_{x 	o 0} f(x)$ must exist and be equal to $f(0)$.Here,$f(0) = k$.We examine the limit $\lim_{x 	o 0} \sin \frac{1}{x}$.As $x 	o 0$,the argument $\frac{1}{x}$ approaches $\infty$ or $-\infty$.The function $\sin \frac{1}{x}$ oscillates infinitely between $-1$ and $1$ as $x$ approaches $0$.Since the limit does not approach a unique finite value,the limit $\lim_{x 	o 0} \sin \frac{1}{x}$ does not exist.Therefore,there is no value of $k$ that can make the function continuous at $x = 0$.

The value of $k$ which makes $f(x) = \begin{cases} \sin \frac{1}{x}, & x \ne 0 \\ k, & x = 0 \end{cases}$ continuous at $x = 0$ is

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