To solve the problem, we need to analyze the function given by:
\[ f(x) = \lim_{n \to \infty} \sin^{2n}(\pi x) + \left[ x + \frac{1}{2} \right] \]
where \([.]\) denotes the greatest integer function. We will check the continuity of \(f(x)\) at the points \(x = 1\) and \(x = \frac{3}{2}\).
### Step 1: Evaluate \(f(1)\)
First, we calculate \(f(1)\):
\[
f(1) = \lim_{n \to \infty} \sin^{2n}(\pi \cdot 1) + \left[ 1 + \frac{1}{2} \right]
\]
Since \(\sin(\pi) = 0\), we have:
\[
\sin^{2n}(\pi) = 0 \quad \text{for any } n \geq 1
\]
Thus,
\[
f(1) = 0 + \left[ 1 + \frac{1}{2} \right] = 0 + [1.5] = 1
\]
### Step 2: Evaluate \(f(1 + h)\)
Next, we consider \(f(1 + h)\) as \(h \to 0\):
\[
f(1 + h) = \lim_{n \to \infty} \sin^{2n}(\pi(1 + h)) + \left[ (1 + h) + \frac{1}{2} \right]
\]
This simplifies to:
\[
f(1 + h) = \lim_{n \to \infty} \sin^{2n}(\pi + \pi h) + \left[ 1 + h + \frac{1}{2} \right]
\]
Since \(\sin(\pi + \pi h) = -\sin(\pi h)\), we have:
\[
\sin^{2n}(\pi + \pi h) = \sin^{2n}(\pi h)
\]
As \(h \to 0\), \(\sin(\pi h) \to \pi h\), and thus:
\[
\sin^{2n}(\pi h) \to 0 \quad \text{(for } n \to \infty \text{)}
\]
Therefore,
\[
f(1 + h) = 0 + \left[ 1 + h + \frac{1}{2} \right] = [1.5 + h]
\]
For \(h\) close to 0, \([1.5 + h] = 1\).
### Step 3: Evaluate \(f(1 - h)\)
Now, we evaluate \(f(1 - h)\):
\[
f(1 - h) = \lim_{n \to \infty} \sin^{2n}(\pi(1 - h)) + \left[ (1 - h) + \frac{1}{2} \right]
\]
This simplifies to:
\[
f(1 - h) = \lim_{n \to \infty} \sin^{2n}(\pi - \pi h) + \left[ 1 - h + \frac{1}{2} \right]
\]
Again, since \(\sin(\pi - \pi h) = \sin(\pi h)\):
\[
f(1 - h) = \lim_{n \to \infty} \sin^{2n}(\pi h) + [1.5 - h]
\]
As before, \(\sin^{2n}(\pi h) \to 0\) as \(h \to 0\):
\[
f(1 - h) = 0 + [1.5 - h]
\]
For \(h\) close to 0, \([1.5 - h] = 1\).
### Step 4: Check Continuity at \(x = 1\)
Now we check the continuity at \(x = 1\):
\[
f(1) = 1, \quad f(1 + h) \to 1, \quad f(1 - h) \to 1 \text{ as } h \to 0
\]
Since \(f(1) = f(1 + h) = f(1 - h)\), the function is continuous at \(x = 1\).
### Step 5: Evaluate \(f\left(\frac{3}{2}\right)\)
Next, we evaluate \(f\left(\frac{3}{2}\right)\):
\[
f\left(\frac{3}{2}\right) = \lim_{n \to \infty} \sin^{2n}\left(\pi \cdot \frac{3}{2}\right) + \left[ \frac{3}{2} + \frac{1}{2} \right]
\]
Since \(\sin\left(\frac{3\pi}{2}\right) = -1\):
\[
\sin^{2n}\left(\frac{3\pi}{2}\right) = 1 \quad \text{(for any } n \geq 1\text{)}
\]
Thus,
\[
f\left(\frac{3}{2}\right) = 1 + [2] = 1 + 2 = 3
\]
### Step 6: Evaluate \(f\left(\frac{3}{2} + h\right)\)
Now, we consider \(f\left(\frac{3}{2} + h\right)\):
\[
f\left(\frac{3}{2} + h\right) = \lim_{n \to \infty} \sin^{2n}\left(\pi\left(\frac{3}{2} + h\right)\right) + \left[ \frac{3}{2} + h + \frac{1}{2} \right]
\]
This simplifies to:
\[
f\left(\frac{3}{2} + h\right) = \lim_{n \to \infty} \sin^{2n}\left(\frac{3\pi}{2} + \pi h\right) + \left[ 3 + h \right]
\]
Since \(\sin\left(\frac{3\pi}{2} + \pi h\right) = -\cos(\pi h)\):
\[
f\left(\frac{3}{2} + h\right) = \lim_{n \to \infty} \sin^{2n}\left(-\cos(\pi h)\right) + [3 + h]
\]
As \(h \to 0\), \(-\cos(\pi h) \to -1\), thus:
\[
\sin^{2n}(-\cos(\pi h)) \to 0 \quad \text{(for } n \to \infty\text{)}
\]
So,
\[
f\left(\frac{3}{2} + h\right) = 0 + [3 + h] = 3
\]
### Step 7: Evaluate \(f\left(\frac{3}{2} - h\right)\)
Now, we evaluate \(f\left(\frac{3}{2} - h\right)\):
\[
f\left(\frac{3}{2} - h\right) = \lim_{n \to \infty} \sin^{2n}\left(\pi\left(\frac{3}{2} - h\right)\right) + \left[ \frac{3}{2} - h + \frac{1}{2} \right]
\]
This simplifies to:
\[
f\left(\frac{3}{2} - h\right) = \lim_{n \to \infty} \sin^{2n}\left(\frac{3\pi}{2} - \pi h\right) + \left[ 3 - h \right]
\]
Since \(\sin\left(\frac{3\pi}{2} - \pi h\right) = -\cos(\pi h)\):
\[
f\left(\frac{3}{2} - h\right) = \lim_{n \to \infty} \sin^{2n}(-\cos(\pi h)) + [3 - h]
\]
As \(h \to 0\), \(-\cos(\pi h) \to -1\), thus:
\[
\sin^{2n}(-\cos(\pi h)) \to 0 \quad \text{(for } n \to \infty\text{)}
\]
So,
\[
f\left(\frac{3}{2} - h\right) = 0 + [3 - h] = 2
\]
### Step 8: Check Continuity at \(x = \frac{3}{2}\)
Now we check the continuity at \(x = \frac{3}{2}\):
\[
f\left(\frac{3}{2}\right) = 3, \quad f\left(\frac{3}{2} + h\right) \to 3, \quad f\left(\frac{3}{2} - h\right) \to 2
\]
Since \(f\left(\frac{3}{2} - h\right) \neq f\left(\frac{3}{2}\right)\), the function is discontinuous at \(x = \frac{3}{2}\).
### Conclusion
The function \(f(x)\) is continuous at \(x = 1\) and discontinuous at \(x = \frac{3}{2}\).
### Final Answer
The function \(f(x)\) is continuous at \(x = 1\) but discontinuous at \(x = \frac{3}{2}\).
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