Let `f (x) =lim _(nto oo) tan ^(-1) (4n ^(2) (1- cos ""(pi)/(n )))and g (x) = lim _(n to oo) (n^(2))/(2) ln cos ((2x)/(n )) ` then `lim _(xto0) (e ^(-2g (x))-ef (x))/(x ^(6))` equals.

Options

A. `8/3`
B. `7/3`
C. `5/3`
D. `2/3`

Answer

To solve the problem, we need to evaluate the limits of the functions \( f(x) \) and \( g(x) \) defined as follows:

1. \( f(x) = \lim_{n \to \infty} \tan^{-1} \left( 4n^2 (1 - \cos(\frac{\pi}{n})) \right) \)
2. \( g(x) = \lim_{n \to \infty} \frac{n^2}{2} \ln \left( \cos\left(\frac{2x}{n}\right) \right) \)

Finally, we need to evaluate:
\[ \lim_{x \to 0} \frac{e^{-2g(x)} - e^{f(x)}}{x^6} \]

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

We start with:
\[ f(x) = \lim_{n \to \infty} \tan^{-1} \left( 4n^2 (1 - \cos(\frac{\pi}{n})) \right) \]

Using the identity \( 1 - \cos(\theta) = 2 \sin^2\left(\frac{\theta}{2}\right) \):
\[ 1 - \cos\left(\frac{\pi}{n}\right) = 2 \sin^2\left(\frac{\pi}{2n}\right) \]

Thus,
\[ f(x) = \lim_{n \to \infty} \tan^{-1} \left( 4n^2 \cdot 2 \sin^2\left(\frac{\pi}{2n}\right) \right) = \lim_{n \to \infty} \tan^{-1} \left( 8n^2 \sin^2\left(\frac{\pi}{2n}\right) \right) \]

Using the small angle approximation \( \sin\left(\theta\right) \approx \theta \) for small \( \theta \):
\[ \sin\left(\frac{\pi}{2n}\right) \approx \frac{\pi}{2n} \]

Thus,
\[ f(x) = \lim_{n \to \infty} \tan^{-1} \left( 8n^2 \left(\frac{\pi}{2n}\right)^2 \right) = \lim_{n \to \infty} \tan^{-1} \left( 8n^2 \cdot \frac{\pi^2}{4n^2} \right) = \lim_{n \to \infty} \tan^{-1} \left( 2\pi^2 \right) = \tan^{-1}(2\pi^2) \]

### Step 2: Evaluate \( g(x) \)

Now we evaluate \( g(x) \):
\[ g(x) = \lim_{n \to \infty} \frac{n^2}{2} \ln \left( \cos\left(\frac{2x}{n}\right) \right) \]

Using the approximation \( \cos(\theta) \approx 1 - \frac{\theta^2}{2} \) for small \( \theta \):
\[ \cos\left(\frac{2x}{n}\right) \approx 1 - \frac{(2x)^2}{2n^2} = 1 - \frac{2x^2}{n^2} \]

Thus,
\[ \ln\left(\cos\left(\frac{2x}{n}\right)\right) \approx \ln\left(1 - \frac{2x^2}{n^2}\right) \approx -\frac{2x^2}{n^2} \]

Therefore,
\[ g(x) = \lim_{n \to \infty} \frac{n^2}{2} \left(-\frac{2x^2}{n^2}\right) = -x^2 \]

### Step 3: Evaluate the final limit

Now we substitute \( f(x) \) and \( g(x) \) into the limit we need to evaluate:
\[ \lim_{x \to 0} \frac{e^{-2g(x)} - e^{f(x)}}{x^6} = \lim_{x \to 0} \frac{e^{2x^2} - e^{\tan^{-1}(2\pi^2)}}{x^6} \]

Using the Taylor expansion for \( e^{2x^2} \):
\[ e^{2x^2} \approx 1 + 2x^2 + \frac{(2x^2)^2}{2} + O(x^6) = 1 + 2x^2 + 2x^4 + O(x^6) \]

Thus,
\[ e^{2x^2} - e^{\tan^{-1}(2\pi^2)} \approx (1 + 2x^2 + 2x^4) - e^{\tan^{-1}(2\pi^2)} = (1 - e^{\tan^{-1}(2\pi^2)}) + 2x^2 + 2x^4 \]

Now, we can evaluate the limit:
\[ \lim_{x \to 0} \frac{(1 - e^{\tan^{-1}(2\pi^2)}) + 2x^2 + 2x^4}{x^6} \]

As \( x \to 0 \), the dominant term is \( 2x^2 \), thus:
\[ \lim_{x \to 0} \frac{2x^2}{x^6} = \lim_{x \to 0} \frac{2}{x^4} \to \infty \]

However, since we are looking for the coefficient of \( x^6 \) in the expansion, we find that the limit evaluates to \( 0 \).

### Final Answer:
\[ \lim_{x \to 0} \frac{e^{-2g(x)} - e^{f(x)}}{x^6} = 0 \]