The value of `lim_(xrarr0)((1+tanx)/(1+sinx))^((2)/(sinx))` is equal to

Options

A. 0
B. 1
C. `-1`
D. None of these

Answer

To solve the limit problem \( \lim_{x \to 0} \left( \frac{1 + \tan x}{1 + \sin x} \right)^{\frac{2}{\sin x}} \), we will follow these steps:

### Step 1: Substitute \( x = 0 \)
First, we will substitute \( x = 0 \) directly into the expression to check if we can evaluate the limit directly.

\[ \frac{1 + \tan(0)}{1 + \sin(0)} = \frac{1 + 0}{1 + 0} = \frac{1}{1} = 1 \]

The exponent \( \frac{2}{\sin(0)} \) becomes \( \frac{2}{0} \), which approaches infinity. Therefore, we have the indeterminate form \( 1^{\infty} \).

### Step 2: Rewrite the limit
Since we have an indeterminate form \( 1^{\infty} \), we can rewrite the limit using the exponential function:

\[ L = \lim_{x \to 0} \left( \frac{1 + \tan x}{1 + \sin x} \right)^{\frac{2}{\sin x}} = e^{\lim_{x \to 0} \frac{2}{\sin x} \left( \frac{1 + \tan x}{1 + \sin x} - 1 \right)} \]

### Step 3: Simplify the expression inside the limit
Now we need to simplify \( \frac{1 + \tan x}{1 + \sin x} - 1 \):

\[ \frac{1 + \tan x}{1 + \sin x} - 1 = \frac{(1 + \tan x) - (1 + \sin x)}{1 + \sin x} = \frac{\tan x - \sin x}{1 + \sin x} \]

### Step 4: Use Taylor series expansions
Next, we can use Taylor series expansions for \( \tan x \) and \( \sin x \) around \( x = 0 \):

- \( \tan x \approx x + \frac{x^3}{3} + O(x^5) \)
- \( \sin x \approx x - \frac{x^3}{6} + O(x^5) \)

Thus,

\[ \tan x - \sin x \approx \left( x + \frac{x^3}{3} \right) - \left( x - \frac{x^3}{6} \right) = \frac{x^3}{3} + \frac{x^3}{6} = \frac{x^3}{2} \]

### Step 5: Substitute back into the limit
Now we substitute back into our limit:

\[ \lim_{x \to 0} \frac{2}{\sin x} \cdot \frac{\tan x - \sin x}{1 + \sin x} = \lim_{x \to 0} \frac{2}{\sin x} \cdot \frac{\frac{x^3}{2}}{1 + \sin x} \]

As \( x \to 0 \), \( \sin x \to x \) and \( 1 + \sin x \to 1 \):

\[ = \lim_{x \to 0} \frac{2 \cdot \frac{x^3}{2}}{x} = \lim_{x \to 0} \frac{x^3}{x} = \lim_{x \to 0} x^2 = 0 \]

### Step 6: Final result
Thus, we have:

\[ L = e^0 = 1 \]

### Conclusion
The value of the limit is:

\[ \boxed{1} \]