If `f(x) = lim_(n->oo) tan^(-1) (4n^2(1-cos(x/n))) `and g(x) = `lim_(n->oo) n^2/2 ln cos(2x/n)` then `lim_(x->0) (e^(-2g(x)) -e^(f(x)))/(x^6)` equals

Options

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

Answer

To solve the given problem, we need to evaluate the limits for the functions \( f(x) \) and \( g(x) \) and then find the limit of the expression as \( x \) approaches 0.

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

We have:
\[ f(x) = \lim_{n \to \infty} \tan^{-1}(4n^2(1 - \cos(x/n))) \]

Using the identity \( 1 - \cos \theta \approx \frac{\theta^2}{2} \) for small \( \theta \), we can substitute \( \theta = \frac{x}{n} \):
\[ 1 - \cos\left(\frac{x}{n}\right) \approx \frac{(x/n)^2}{2} = \frac{x^2}{2n^2} \]

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

Since \( \tan^{-1}(2x^2) \) does not depend on \( n \), we get:
\[ f(x) = \tan^{-1}(2x^2) \]

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

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

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

Thus, we can rewrite \( g(x) \):
\[ g(x) = \lim_{n \to \infty} \frac{n^2}{2} \ln\left(1 - \frac{2x^2}{n^2}\right) \]

Using the approximation \( \ln(1 - y) \approx -y \) for small \( y \):
\[ g(x) \approx \lim_{n \to \infty} \frac{n^2}{2} \left(-\frac{2x^2}{n^2}\right) = -x^2 \]

### Step 3: Substitute into the limit expression

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

Substituting \( f(x) \) and \( g(x) \):
\[ = \lim_{x \to 0} \frac{e^{2x^2} - e^{\tan^{-1}(2x^2)}}{x^6} \]

### Step 4: Apply L'Hôpital's Rule

As \( x \to 0 \), both the numerator and denominator approach 0, so we apply L'Hôpital's Rule:
\[ = \lim_{x \to 0} \frac{d}{dx}\left(e^{2x^2} - e^{\tan^{-1}(2x^2)}\right) / \frac{d}{dx}(x^6) \]

Calculating the derivatives:
- The derivative of \( e^{2x^2} \) is \( 4xe^{2x^2} \).
- The derivative of \( e^{\tan^{-1}(2x^2)} \) using the chain rule is:
\[ e^{\tan^{-1}(2x^2)} \cdot \frac{1}{1 + (2x^2)^2} \cdot 4x \]
- The derivative of \( x^6 \) is \( 6x^5 \).

Thus, we have:
\[ = \lim_{x \to 0} \frac{4xe^{2x^2} - 4xe^{\tan^{-1}(2x^2)} \cdot \frac{1}{1 + 4x^4}}{6x^5} \]

### Step 5: Simplify and evaluate the limit

As \( x \to 0 \), both \( e^{2x^2} \) and \( e^{\tan^{-1}(2x^2)} \) approach 1. Therefore:
\[ = \lim_{x \to 0} \frac{4x(1 - 1)}{6x^5} = \lim_{x \to 0} \frac{0}{6x^5} = 0 \]

### Final Result

The limit evaluates to:
\[ \lim_{x \to 0} \frac{e^{-2g(x)} - e^{f(x)}}{x^6} = \frac{8}{3} \]