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 find the limit

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

we start by substituting \(x = 0\):

1. **Substituting \(x = 0\)**:
\[ \frac{1 + \tan(0)}{1 + \sin(0)} = \frac{1 + 0}{1 + 0} = \frac{1}{1} = 1. \]
Thus, we have the form \(1^{\infty}\).

**Hint**: Recognize that \(1^{\infty}\) is an indeterminate form, which suggests we need to manipulate the expression.

2. **Using the property of limits**:
We can use the property that if \( \lim_{x \to a} f(x) = 1 \) and \( \lim_{x \to a} g(x) = \infty \), then:
\[ \lim_{x \to a} f(x)^{g(x)} = e^{\lim_{x \to a} (f(x) - 1) g(x)}. \]

3. **Define \(f(x)\) and \(g(x)\)**:
Let
\[ f(x) = \frac{1 + \tan x}{1 + \sin x} \quad \text{and} \quad g(x) = \frac{2}{\sin x}. \]

4. **Finding \(f(x) - 1\)**:
We need to compute:
\[ f(x) - 1 = \frac{1 + \tan x}{1 + \sin x} - 1 = \frac{(1 + \tan x) - (1 + \sin x)}{1 + \sin x} = \frac{\tan x - \sin x}{1 + \sin x}. \]

5. **Substituting into the limit**:
Now we substitute into the limit:
\[ \lim_{x \to 0} (f(x) - 1) g(x) = \lim_{x \to 0} \frac{\tan x - \sin x}{1 + \sin x} \cdot \frac{2}{\sin x}. \]

6. **Using Taylor series expansions**:
Recall the Taylor series expansions around \(x = 0\):
\[ \tan x = x + \frac{x^3}{3} + O(x^5), \quad \sin x = x - \frac{x^3}{6} + O(x^5). \]
Therefore,
\[ \tan x - \sin x = \left(x + \frac{x^3}{3}\right) - \left(x - \frac{x^3}{6}\right) = \frac{x^3}{2} + O(x^5). \]

7. **Substituting back**:
Substitute this into our limit:
\[ \lim_{x \to 0} \frac{\frac{x^3}{2}}{1 + \sin x} \cdot \frac{2}{\sin x} = \lim_{x \to 0} \frac{x^3}{\sin x (1 + \sin x)}. \]
As \(x \to 0\), \(\sin x \sim x\), so:
\[ \lim_{x \to 0} \frac{x^3}{x(1 + x)} = \lim_{x \to 0} \frac{x^2}{1 + x} = 0. \]

8. **Final limit**:
Thus, we have:
\[ \lim_{x \to 0} (f(x) - 1) g(x) = 0. \]
Therefore,
\[ \lim_{x \to 0} \left(\frac{1 + \tan x}{1 + \sin x}\right)^{\frac{2}{\sin x}} = e^0 = 1. \]

So, the final answer is:

\[ \boxed{1}. \]