If `f(x+ y) = f(x) + f(y)` for `x, y in R` and `f(1) = 1`, then find the value of `lim_(x->0)(2^(f(tan x)-2^f(sin x)))/(x^2*f(sin x))`

Options

A. `log2`
B. `log4`
C. `logsqrt(2)`
D. `log8`

Answer

To solve the problem, we need to find the limit:

\[ \lim_{x \to 0} \frac{2^{f(\tan x)} - 2^{f(\sin x)}}{x^2 f(\sin x)} \]

Given that \( f(x+y) = f(x) + f(y) \) for all \( x, y \in \mathbb{R} \) and \( f(1) = 1 \), we can deduce that \( f(x) = x \) for all \( x \in \mathbb{R} \).

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

Using the functional equation \( f(x+y) = f(x) + f(y) \):
- Let \( x = 0 \) and \( y = 0 \):
\[ f(0) = f(0) + f(0) \implies f(0) = 0 \]
- Let \( x = 1 \) and \( y = 1 \):
\[ f(2) = f(1) + f(1) = 1 + 1 = 2 \]
- By induction or by checking integer values, we can conclude that \( f(n) = n \) for all integers \( n \). Since \( f \) is linear, we can generalize that \( f(x) = x \) for all \( x \in \mathbb{R} \).

### Step 2: Substitute \( f(x) \) into the limit

Now substituting \( f(\tan x) \) and \( f(\sin x) \):
\[ \lim_{x \to 0} \frac{2^{\tan x} - 2^{\sin x}}{x^2 \sin x} \]

### Step 3: Evaluate the limit

As \( x \to 0 \):
- \( \tan x \to 0 \) and \( \sin x \to 0 \), thus \( 2^{\tan x} \to 1 \) and \( 2^{\sin x} \to 1 \).
- The limit takes the indeterminate form \( \frac{0}{0} \).

### Step 4: Simplify the expression

We can rewrite the limit:
\[ \lim_{x \to 0} \frac{2^{\tan x} - 1 - (2^{\sin x} - 1)}{x^2 \sin x} \]

Using the Taylor expansion for \( 2^u \) around \( u = 0 \):
\[ 2^u \approx 1 + u \ln 2 \quad \text{for small } u \]

Thus:
\[ 2^{\tan x} - 1 \approx \tan x \ln 2 \quad \text{and} \quad 2^{\sin x} - 1 \approx \sin x \ln 2 \]

Substituting these approximations:
\[ \lim_{x \to 0} \frac{(\tan x - \sin x) \ln 2}{x^2 \sin x} \]

### Step 5: Use the identity for \( \tan x - \sin x \)

Using the identity \( \tan x = \frac{\sin x}{\cos x} \):
\[ \tan x - \sin x = \frac{\sin x}{\cos x} - \sin x = \sin x \left(\frac{1}{\cos x} - 1\right) = \sin x \left(\frac{1 - \cos x}{\cos x}\right) \]

### Step 6: Substitute back into the limit

Now we have:
\[ \lim_{x \to 0} \frac{\sin x \cdot \frac{1 - \cos x}{\cos x} \ln 2}{x^2 \sin x} = \lim_{x \to 0} \frac{(1 - \cos x) \ln 2}{x^2 \cos x} \]

### Step 7: Evaluate the limit

Using the fact that \( 1 - \cos x \approx \frac{x^2}{2} \) as \( x \to 0 \):
\[ \lim_{x \to 0} \frac{\frac{x^2}{2} \ln 2}{x^2 \cos x} = \lim_{x \to 0} \frac{\frac{1}{2} \ln 2}{\cos x} = \frac{1}{2} \ln 2 \]

### Final Answer

Thus, the final result is:
\[ \frac{1}{2} \ln 2 \]