A function is defined as `f(x)=lim_(x->oo)[cos^(2n) x` , if `x<0)` , `nsqrt(sqrt(1+x^n,` if `0<=x,+1)` , `1/(1+x^n),` if `x>1` which of the following does not hold good?

Options

A. continuous at x=0 but discontinuous at x=1
B. continuous at x=1 but discontinuous at x=0
C. continuous both as x=1 and x=0
D. discontinuous both at x=1 and x=0

Answer

To solve the problem, we need to analyze the function defined piecewise and check for continuity at the critical points \( x = 0 \) and \( x = 1 \).

The function \( f(x) \) is defined as follows:

1. \( f(x) = \lim_{n \to \infty} [\cos^{2n}(x)] \) for \( x < 0 \)
2. \( f(x) = \sqrt[n]{\sqrt{1 + x^n}} \) for \( 0 \leq x \leq 1 \)
3. \( f(x) = \frac{1}{1 + x^n} \) for \( x > 1 \)

We need to determine which of the statements regarding the continuity of \( f(x) \) does not hold.

### Step 1: Check Continuity at \( x = 0 \)

**Left-Hand Limit (LHL) at \( x = 0 \):**
For \( x < 0 \):
\[ f(0^-) = \lim_{n \to \infty} \cos^{2n}(0) = \lim_{n \to \infty} 1^{2n} = 1 \]
However, since we are approaching from the left, we need to consider values slightly less than zero. The maximum value of \( \cos^2(x) \) is 1, but for \( x \) slightly less than 0, \( \cos^2(x) < 1 \). Thus:
\[ f(0^-) = \lim_{n \to \infty} \cos^{2n}(x) \to 0 \text{ as } n \to \infty \]

**Right-Hand Limit (RHL) at \( x = 0 \):**
For \( 0 \leq x \leq 1 \):
\[ f(0^+) = \lim_{n \to \infty} \sqrt[n]{\sqrt{1 + 0^n}} = \sqrt[n]{\sqrt{1}} = 1 \]

Since:
\[ \text{LHL} = 0 \quad \text{and} \quad \text{RHL} = 1 \]
Thus, \( f(x) \) is discontinuous at \( x = 0 \).

### Step 2: Check Continuity at \( x = 1 \)

**Left-Hand Limit (LHL) at \( x = 1 \):**
For \( x < 1 \):
\[ f(1^-) = \lim_{n \to \infty} \sqrt[n]{\sqrt{1 + 1^n}} = \lim_{n \to \infty} \sqrt[n]{\sqrt{2}} = 2^{1/n} \to 1 \text{ as } n \to \infty \]

**Right-Hand Limit (RHL) at \( x = 1 \):**
For \( x > 1 \):
\[ f(1^+) = \lim_{n \to \infty} \frac{1}{1 + 1^n} = \frac{1}{1 + 1} = \frac{1}{2} \]

Since:
\[ \text{LHL} = 1 \quad \text{and} \quad \text{RHL} = \frac{1}{2} \]
Thus, \( f(x) \) is also discontinuous at \( x = 1 \).

### Conclusion

The function \( f(x) \) is discontinuous at both \( x = 0 \) and \( x = 1 \). Therefore, any statement claiming that the function is continuous at either of these points is false.

### Final Answer

The statement that does not hold good is that the function is continuous at \( x = 0 \) or \( x = 1 \).

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