`f(x) = lim_(n rarr oo) ((x-1)^(2n) - 1)/((x-1)^(2n) + 1)` is discontinuous at (A) x=0 only (B) x=2 only (C) x=0 and 2 (D) none of these

Options

A. x= 0 only
B. x=2 only
C. x= 0 and 2
D. none of these

Answer

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

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

and determine where it is discontinuous, specifically at \(x = 0\) and \(x = 2\).

### Step 1: Evaluate \(f(0)\)

Substituting \(x = 0\):

\[ f(0) = \lim_{n \to \infty} \frac{(0-1)^{2n} - 1}{(0-1)^{2n} + 1} = \lim_{n \to \infty} \frac{(-1)^{2n} - 1}{(-1)^{2n} + 1} \]

Since \((-1)^{2n} = 1\):

\[ f(0) = \lim_{n \to \infty} \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0 \]

### Step 2: Evaluate the right-hand limit as \(x \to 0^+\)

For \(x\) approaching \(0\) from the right (\(x \to 0^+\)):

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

Here, \(x - 1 < 0\) (since \(x < 1\)), thus \((x-1)^{2n} \to 1\) as \(n \to \infty\):

\[ f(x) = \lim_{n \to \infty} \frac{(-1)^{2n} - 1}{(-1)^{2n} + 1} = \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0 \]

### Step 3: Evaluate the left-hand limit as \(x \to 0^-\)

For \(x\) approaching \(0\) from the left (\(x \to 0^-\)):

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

Here, \(x - 1 < 0\) and \((x-1)^{2n} \to 1\) as \(n \to \infty\):

\[ f(x) = \lim_{n \to \infty} \frac{(-1)^{2n} - 1}{(-1)^{2n} + 1} = \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0 \]

### Step 4: Check for discontinuity at \(x = 0\)

Since \(f(0) = 0\) and both the left-hand and right-hand limits as \(x \to 0\) are equal to \(0\), the function is continuous at \(x = 0\).

### Step 5: Evaluate \(f(2)\)

Substituting \(x = 2\):

\[ f(2) = \lim_{n \to \infty} \frac{(2-1)^{2n} - 1}{(2-1)^{2n} + 1} = \lim_{n \to \infty} \frac{1^{2n} - 1}{1^{2n} + 1} = \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0 \]

### Step 6: Evaluate the right-hand limit as \(x \to 2^+\)

For \(x\) approaching \(2\) from the right (\(x \to 2^+\)):

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

Here, \(x - 1 > 1\) (since \(x > 2\)), thus \((x-1)^{2n} \to \infty\):

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

### Step 7: Check for discontinuity at \(x = 2\)

Since \(f(2) = 0\) and the right-hand limit as \(x \to 2^+\) is \(1\), the function is discontinuous at \(x = 2\).

### Conclusion

The function \(f(x)\) is discontinuous at \(x = 2\) but continuous at \(x = 0\). Therefore, the answer is:

**(B) x = 2 only.**