Question

If \[ f(x)=\left(\frac{1+\sin x}{1-\sin x}\right)^{\tan x}, \] then find \[ \lim_{x\to0}\frac{\ln f(x)}{x^2}. \]

• A. \(1\)
• B. \(2\)
• C. \(4\)
• D. \(0\)

Hint:

For difficult exponential limits, always convert powers into logarithmic form before applying series expansions.

Answer 1

The Correct Option is B

Concept: Very high-level limits involving powers are handled using logarithmic transformation and standard series approximations.

Step 1: Taking logarithm.
Given: \[ f(x)=\left(\frac{1+\sin x}{1-\sin x}\right)^{\tan x} \] Taking logarithm: \[ \ln f(x) = \tan x\cdot \ln\left(\frac{1+\sin x}{1-\sin x}\right) \] Hence, \[ \frac{\ln f(x)}{x^2} = \frac{\tan x}{x} \cdot \frac{1}{x} \ln\left(\frac{1+\sin x}{1-\sin x}\right) \]

Step 2: Using standard expansions.
As \(x\to0\), \[ \tan x\sim x \] Thus, \[ \frac{\tan x}{x}\to1 \] Also, \[ \ln\left(\frac{1+t}{1-t}\right) = 2\left(t+\frac{t^3}{3}+\cdots\right) \] Putting \(t=\sin x\), \[ \ln\left(\frac{1+\sin x}{1-\sin x}\right) \sim2\sin x \] Since \[ \sin x\sim x \] we get: \[ \ln\left(\frac{1+\sin x}{1-\sin x}\right)\sim2x \] Therefore, \[ \frac{1}{x} \ln\left(\frac{1+\sin x}{1-\sin x}\right)\to2 \]

Step 3: Evaluating the limit.
Hence, \[ \lim_{x\to0}\frac{\ln f(x)}{x^2} = 1\times2 \] \[ =2 \]