If f(x) = cos left[ frac{pi}{x} right] cos le | vedclass

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(D) The function is $f(x) = \cos \left[ \frac{\pi}{x} \right] \cos \left( \frac{\pi}{2} (x - 1) \right)$.At $x = 0$,the function is undefined because

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$x = 1$ and $x = 2$

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(D) The function is $f(x) = \cos \left[ \frac{\pi}{x} ight] \cos \left( \frac{\pi}{2} (x - 1) ight)$.At $x = 0$,the function is undefined because the denominator in $\frac{\pi}{x}$ is zero.At $x = 1$,$f(1) = \cos \left[ \frac{\pi}{1} ight] \cos \left( \frac{\pi}{2} (1 - 1) ight) = \cos(3) \cos(0) = \cos(3)$. The limit as $x 	o 1$ exists and equals $f(1)$,so it is continuous at $x = 1$.At $x = 2$,$f(2) = \cos \left[ \frac{\pi}{2} ight] \cos \left( \frac{\pi}{2} (2 - 1) ight) = \cos(1) \cos \left( \frac{\pi}{2} ight) = \cos(1) 	imes 0 = 0$. The limit as $x 	o 2$ exists and equals $f(2)$,so it is continuous at $x = 2$.Therefore,the function is continuous at both $x = 1$ and $x = 2$.

If $f(x) = \cos \left[ \frac{\pi}{x} \right] \cos \left( \frac{\pi}{2} (x - 1) \right)$,then $f(x)$ is continuous at: (where $[x]$ is the greatest integer function of $x$)

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