Question

(c) ( f(x) = llbracket sin(x)
rbracket ), ( a = pi ) (recall that ( llbracket h(x)
rbracket ) means the largest integer that is less than or equal to ( h(x) ).)
( \bigcirc ) the discontinuity is removable.
( \bigcirc ) the discontinuity is not removable.
( g(x) = ) blank with i

Explanation:

Step1: Analyze \( \sin(\pi) \)

We know that \( \sin(\pi) = 0 \). So \( f(\pi)=\llbracket\sin(\pi) rbracket=\llbracket0 rbracket = 0 \).

Step2: Analyze left - hand limit as \( x

ightarrow\pi^{-} \)
As \( x ightarrow\pi^{-} \) (approaching \( \pi \) from the left), \( \sin(x) \) is in the interval \( (0, 1) \) (because \( \sin(x) \) is positive and less than 1 when \( x \) is in \( (0,\pi) \) and approaching \( \pi \)). The greatest integer less than or equal to a number in \( (0, 1) \) is 0? Wait, no. Wait, when \( x \) is in \( (\pi/2,\pi) \), \( \sin(x) \) is in \( (0, 1) \)? No, when \( x=\pi/2 \), \( \sin(x) = 1 \), when \( x \) approaches \( \pi \) from the left, \( \sin(x) \) approaches 0 from the positive side. Wait, actually, for \( x\in(\pi - \epsilon,\pi) \) where \( \epsilon>0 \) is small, \( \sin(x)=\sin(\pi-( \pi - x))=\sin(\pi - x) \), and \( \pi - x\in(0,\epsilon) \), so \( \sin(x)\in(0,\sin(\epsilon)) \). But when we take the greatest integer function, \( \llbracket\sin(x) rbracket \) for \( \sin(x)\in(0, 1) \) is 0? Wait, no, wait. Wait, if \( x \) is in \( (\pi/2,\pi) \), \( \sin(x) \) is in \( (0, 1) \)? No, \( \sin(\pi/2) = 1 \), \( \sin(\pi)=0 \), so for \( x\in(\pi/2,\pi) \), \( \sin(x)\in(0, 1) \), so \( \llbracket\sin(x) rbracket = 0 \)? Wait, no, wait, the greatest integer less than or equal to a number between 0 and 1 (not including 1) is 0. But when \( x=\pi/2 \), \( \sin(x) = 1 \), so \( \llbracket\sin(\pi/2) rbracket=\llbracket1 rbracket = 1 \). Now, let's consider the right - hand limit as \( x ightarrow\pi^{+} \). As \( x ightarrow\pi^{+} \), \( \sin(x)=\sin(\pi+(x - \pi))=-\sin(x - \pi) \), and \( x-\pi\in(0,\epsilon) \) for small \( \epsilon>0 \), so \( \sin(x)\in(-\sin(\epsilon),0) \). The greatest integer less than or equal to a number in \( (-1,0) \) is - 1.
So, \( \lim_{x ightarrow\pi^{-}}f(x)=\lim_{x ightarrow\pi^{-}}\llbracket\sin(x) rbracket = 0 \) (since \( \sin(x)\in(0, 1) \) when \( x ightarrow\pi^{-} \)), \( f(\pi)=\llbracket\sin(\pi) rbracket = 0 \), and \( \lim_{x ightarrow\pi^{+}}f(x)=\lim_{x ightarrow\pi^{+}}\llbracket\sin(x) rbracket=- 1 \) (since \( \sin(x)\in(-1,0) \) when \( x ightarrow\pi^{+} \)). Since \( \lim_{x ightarrow\pi^{+}}f(x) eq\lim_{x ightarrow\pi^{-}}f(x) \), the limit as \( x ightarrow\pi \) does not exist. A removable discontinuity requires that \( \lim_{x ightarrow a}f(x) \) exists but is not equal to \( f(a) \) (or \( f(a) \) is not defined). Since the limit does not exist here, the discontinuity is not removable.
For the function \( g(x) \), if we want to check removable discontinuity, but in this case, since the discontinuity is not removable, we can't re - define \( g(x) \) to make it continuous at \( x = \pi \) because the left - hand and right - hand limits are not equal. But maybe there is a mistake in my earlier analysis. Wait, let's re - examine \( \sin(x) \) around \( x=\pi \).
When \( x ightarrow\pi^{-} \), \( x=\pi-\delta \), \( \delta ightarrow0^{+} \), \( \sin(x)=\sin(\pi - \delta)=\sin\delta ightarrow0^{+} \), so \( \llbracket\sin(x) rbracket=\llbracket\sin\delta rbracket \). Since \( \sin\delta\in(0, 1) \) for \( \delta\in(0,\pi/2) \), \( \llbracket\sin\delta rbracket = 0 \). When \( x ightarrow\pi^{+} \), \( x=\pi+\delta \), \( \delta ightarrow0^{+} \), \( \sin(x)=\sin(\pi+\delta)=-\sin\delta\in(-1,0) \), so \( \llbracket\sin(x) rbracket=\llbracket-\sin\delta rbracket=- 1 \). And \( f(\pi)=\llbracket\sin(\pi) rbracket=\llbracket0 rbracket = 0 \). So \( \lim_{x ightarrow\pi^{-}}f(x) = 0 \), \( \lim_{x ightarrow\pi^{+}}f(x)=-1 \), so \( \lim_{x ightarrow\pi}f(x) \) does not exist. Therefore, the discontinuity is not removable.

Answer:

The discontinuity is not removable. For \( g(x) \), since the discontinuity is not removable, there is no function \( g(x) \) (that is a re - definition of \( f(x) \) at \( x = \pi \) and equal to \( f(x) \) elsewhere) that makes \( g(x) \) continuous at \( x=\pi \) because the left and right limits are not equal. But based on the question's multiple - choice part, the answer for the discontinuity type is "The discontinuity is not removable."