Let `f:(-pi/2, pi/2)->R`, `f(x) ={lim_(n->oo) ((tanx)^(2n)+x^2)/(sin^2x+(tanx)^(2n)) x in 0 , n in N`; `1` if `x=0` which of the following holds good? (a) `f(-(pi^(-))/4)=f((pi^(+))/4)` (b) `f(-(pi^(-))/4)=f(-(pi^(+))/4)` (c) `f((pi^(-))/4)=f((pi^(+))/4)` (d) `f(0^(+))=f(0)=f(0^(-))`

Options

A. `f(-(pi^(-))/4)=f((pi^(+))/4)`
B. `f(-(pi^(-))/4)=f(-(pi^(+))/4)`
C. `f((pi^(-))/4)=f((pi^(+))/4)`
D. `f(0^(+))=f(0)=f(0^(-))`

Answer

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

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

for \( x \in (-\frac{\pi}{2}, \frac{\pi}{2}) \) and \( f(0) = 1 \).

### Step 1: Analyze the limit for \( x = 0 \)

When \( x = 0 \):

\[ f(0) = 1 \]

This is given in the problem statement.

### Step 2: Analyze the limit for \( x = \frac{\pi}{4} \)

Next, we evaluate \( f(\frac{\pi}{4}) \):

\[ f\left(\frac{\pi}{4}\right) = \lim_{n \to \infty} \frac{(\tan(\frac{\pi}{4}))^{2n} + \left(\frac{\pi}{4}\right)^2}{\sin^2(\frac{\pi}{4}) + (\tan(\frac{\pi}{4}))^{2n}} \]

Since \( \tan(\frac{\pi}{4}) = 1 \) and \( \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} \):

\[ f\left(\frac{\pi}{4}\right) = \lim_{n \to \infty} \frac{1^{2n} + \left(\frac{\pi}{4}\right)^2}{\left(\frac{1}{\sqrt{2}}\right)^2 + 1^{2n}} = \lim_{n \to \infty} \frac{1 + \frac{\pi^2}{16}}{\frac{1}{2} + 1} \]

Calculating the denominator:

\[ \frac{1}{2} + 1 = \frac{3}{2} \]

Thus, we have:

\[ f\left(\frac{\pi}{4}\right) = \frac{1 + \frac{\pi^2}{16}}{\frac{3}{2}} = \frac{2(1 + \frac{\pi^2}{16})}{3} = \frac{2 + \frac{\pi^2}{8}}{3} \]

### Step 3: Analyze the limit for \( x = -\frac{\pi}{4} \)

Now, we evaluate \( f(-\frac{\pi}{4}) \):

\[ f\left(-\frac{\pi}{4}\right) = \lim_{n \to \infty} \frac{(\tan(-\frac{\pi}{4}))^{2n} + \left(-\frac{\pi}{4}\right)^2}{\sin^2(-\frac{\pi}{4}) + (\tan(-\frac{\pi}{4}))^{2n}} \]

Since \( \tan(-\frac{\pi}{4}) = -1 \) and \( \sin(-\frac{\pi}{4}) = -\frac{1}{\sqrt{2}} \):

\[ f\left(-\frac{\pi}{4}\right) = \lim_{n \to \infty} \frac{(-1)^{2n} + \left(\frac{\pi}{4}\right)^2}{\left(-\frac{1}{\sqrt{2}}\right)^2 + (-1)^{2n}} = \lim_{n \to \infty} \frac{1 + \frac{\pi^2}{16}}{\frac{1}{2} + 1} \]

This is the same as \( f(\frac{\pi}{4}) \):

\[ f\left(-\frac{\pi}{4}\right) = \frac{2 + \frac{\pi^2}{8}}{3} \]

### Step 4: Conclusion for options

1. \( f(-\frac{\pi}{4}) = f(\frac{\pi}{4}) \) holds true.
2. \( f(0^+) = f(0) = f(0^-) = 1 \) also holds true.

Thus, the correct options are:

- (a) \( f(-\frac{\pi}{4}) = f(\frac{\pi}{4}) \)
- (d) \( f(0^+) = f(0) = f(0^-) \)