Step 1 : Understanding the Question:
The topic for this problem is Limits, specifically focusing on Indeterminate Forms of the type $1^\infty$. The limit involves a composite function where both the base and the exponent change as $x \rightarrow 0$. By taking the natural logarithm, we convert the exponential limit into a product, which can then be solved using standard trigonometric limits or series expansions.
Step 2 : Key Formulas and approach:
1. Power to Multiplier: $\ln(a^b) = b \ln a$.
2. Logarithmic Series: $\ln(1+t) = t - \frac{t^2}{2} + \dots$ for small $t$.
3. Standard Limits: $\lim_{x\to 0} \frac{\sin x}{x} = 1$ and $\lim_{x\to 0} \frac{\tan x}{x} = 1$.
4. Approach: Take the natural log of $f(x)$, express the function as a product, use small-angle approximations for $\sin x$ and $\tan x$, and evaluate the limit as $x \rightarrow 0$.
Step 3 : Detailed Explanation:
We start by taking the natural log of $f(x)$: $\ln f(x) = \tan x \cdot \ln \left(\frac{1+\sin x}{1-\sin x}\right)$.
We need to evaluate $L = \lim_{x\to 0} \frac{\tan x \cdot \ln \left(\frac{1+\sin x}{1-\sin x}\right)}{x^2}$.
We can split this into two parts: $L = \left( \lim_{x\to 0} \frac{\tan x}{x} \right) \cdot \left( \lim_{x\to 0} \frac{\ln(1+\sin x) - \ln(1-\sin x)}{x} \right)$.
The first part is a standard limit: $\lim_{x\to 0} \frac{\tan x}{x} = 1$.
For the second part, use the approximation $\ln(1+t) \approx t$ for very small $t$. As $x \rightarrow 0$, $\sin x$ also goes to 0.
So, $\ln(1+\sin x) \approx \sin x$ and $\ln(1-\sin x) \approx -\sin x$.
The second limit becomes: $\lim_{x\to 0} \frac{\sin x - (-\sin x)}{x} = \lim_{x\to 0} \frac{2\sin x}{x}$.
Using the standard limit $\frac{\sin x}{x} = 1$, this evaluates to $2(1) = 2$.
Multiplying both parts together: $L = 1 \times 2 = 2$.
Step 4 : Final Answer:
By converting the function to its logarithmic form and applying standard limit approximations, we find the limit to be 2. The correct option is (B).