The function `f(x)=lim_(nrarroo)((x-2)^(2n)-1)/((x-2)^(2n)+1) (AA n in N)` is discontinuous at

Options

A. x = 1 only
B. x = 3 only
C. x = 1 and 3
D. x = 0, 1 and 2

Answer

To solve the problem, we need to analyze the function given by:

\[ f(x) = \lim_{n \to \infty} \frac{(x-2)^{2n} - 1}{(x-2)^{2n} + 1} \]

### Step 1: Rewrite the function
We can rewrite the function by factoring out \((x-2)^{2n}\):

\[ f(x) = \lim_{n \to \infty} \frac{(x-2)^{2n}(1 - \frac{1}{(x-2)^{2n}})}{(x-2)^{2n}(1 + \frac{1}{(x-2)^{2n}})} = \lim_{n \to \infty} \frac{1 - \frac{1}{(x-2)^{2n}}}{1 + \frac{1}{(x-2)^{2n}}} \]

### Step 2: Analyze the limit based on the value of \(x\)
Now we need to consider different cases based on the value of \(x\):

1. **Case 1: \(x - 2 = 0\) (i.e., \(x = 2\))**
- Here, both the numerator and denominator approach \(0\):
\[ f(2) = \lim_{n \to \infty} \frac{0 - 1}{0 + 1} = -1 \]

2. **Case 2: \(x - 2 > 0\) (i.e., \(x > 2\))**
- As \(n\) approaches infinity, \((x-2)^{2n} \to \infty\):
\[ f(x) = \lim_{n \to \infty} \frac{1 - 0}{1 + 0} = 1 \]

3. **Case 3: \(x - 2 < 0\) (i.e., \(x < 2\))**
- As \(n\) approaches infinity, \((x-2)^{2n} \to 0\):
\[ f(x) = \lim_{n \to \infty} \frac{0 - 1}{0 + 1} = -1 \]

### Step 3: Summarize the results
From the above analysis, we can summarize the function \(f(x)\) as follows:

- \(f(x) = -1\) for \(x < 2\)
- \(f(x) = -1\) for \(x = 2\)
- \(f(x) = 1\) for \(x > 2\)

### Step 4: Identify points of discontinuity
Now we check for points of discontinuity:

1. At \(x = 2\):
- Left-hand limit as \(x \to 2^{-}\) is \(-1\)
- Right-hand limit as \(x \to 2^{+}\) is \(1\)
- Since the left-hand limit and right-hand limit are not equal, \(f(x)\) is discontinuous at \(x = 2\).

2. For \(x = 1\) and \(x = 3\):
- At \(x = 1\):
- Left-hand limit is \(-1\)
- Right-hand limit is \(-1\)
- The function is continuous at \(x = 1\).
- At \(x = 3\):
- Left-hand limit is \(1\)
- Right-hand limit is \(1\)
- The function is continuous at \(x = 3\).

### Conclusion
The function \(f(x)\) is discontinuous at \(x = 2\).