If `phi(x)=lim_(n->oo)(x^(2n)(f(x)+g(x)))/(1+x^(2n))` then which of the following is correct

Options

A. `phi (x)=g(x) "for all"x in R`
B. `phi (x)=f(x)"for all" x in R `
C. `phix={(g(x)"for" -1ltxlt1,,),(f(x)"for"|x|ge1,,):}`
D. `phix={(g(x)"for" |x|lt1,,),(f(x)"for"|x|gt1,,),((f(x)+g(x))/(2)"for"|x|=1,,):}`

Explanation

To solve the problem, we need to analyze the limit given in the function \( \phi(x) \):

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

We will consider three cases based on the value of \( |x| \).

### Step 1: Case 1 - When \( |x| < 1 \)

If \( |x| < 1 \), as \( n \to \infty \), \( x^{2n} \) approaches 0. Therefore, we can simplify \( \phi(x) \):

\[ \phi(x) = \lim_{n \to \infty} \frac{0 \cdot (f(x) + g(x))}{1 + 0} = 0 \]

### Step 2: Case 2 - When \( |x| > 1 \)

If \( |x| > 1 \), as \( n \to \infty \), \( x^{2n} \) approaches infinity. Thus, we can simplify \( \phi(x) \):

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

### Step 3: Case 3 - When \( |x| = 1 \)

If \( |x| = 1 \), then \( x^{2n} = 1 \) for all \( n \). Hence, we have:

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

### Conclusion

Now we can summarize the results from the three cases:

1. If \( |x| < 1 \), then \( \phi(x) = 0 \).
2. If \( |x| > 1 \), then \( \phi(x) = f(x) + g(x) \).
3. If \( |x| = 1 \), then \( \phi(x) = \frac{f(x) + g(x)}{2} \).

Based on these results, we can conclude that the correct option is:

- \( \phi(x) = 0 \) for \( |x| < 1 \)
- \( \phi(x) = f(x) + g(x) \) for \( |x| > 1 \)
- \( \phi(x) = \frac{f(x) + g(x)}{2} \) for \( |x| = 1 \)