The value of `f(0)` such that `f(x)=((1+tanx)/(1+sinx))^(cosecx)` can be made continuous at `x=0` is

Options

A. 1
B. e
C. -1
D. `e^(-1)`

Answer

To find the value of \( f(0) \) such that the function

\[ f(x) = \left( \frac{1 + \tan x}{1 + \sin x} \right)^{\csc x} \]

is continuous at \( x = 0 \), we need to evaluate the limit of \( f(x) \) as \( x \) approaches 0.

### Step 1: Evaluate the limit as \( x \to 0 \)

We start by finding:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \left( \frac{1 + \tan x}{1 + \sin x} \right)^{\csc x} \]

### Step 2: Identify the indeterminate form

Substituting \( x = 0 \):

- \( \tan(0) = 0 \) so \( 1 + \tan(0) = 1 \)
- \( \sin(0) = 0 \) so \( 1 + \sin(0) = 1 \)

Thus, we have:

\[ f(0) = \left( \frac{1 + 0}{1 + 0} \right)^{\csc(0)} = 1^{\infty} \]

This is an indeterminate form \( 1^{\infty} \).

### Step 3: Apply the limit formula for \( 1^{\infty} \)

To resolve this, we can use the limit:

\[ \lim_{x \to a} f(x)^{g(x)} = e^{\lim_{x \to a} g(x) \cdot (f(x) - 1)} \]

where \( f(x) = \frac{1 + \tan x}{1 + \sin x} \) and \( g(x) = \csc x \).

### Step 4: Rewrite the limit expression

We rewrite the limit:

\[ \lim_{x \to 0} \csc x \cdot \left( \frac{1 + \tan x}{1 + \sin x} - 1 \right) \]

### Step 5: Simplify the expression inside the limit

We simplify:

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

### Step 6: Rewrite \( \tan x \) in terms of \( \sin x \)

Using \( \tan x = \frac{\sin x}{\cos x} \):

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

### Step 7: Substitute back into the limit

Now we have:

\[ \lim_{x \to 0} \csc x \cdot \frac{\sin x \left( \frac{1 - \cos x}{\cos x} \right)}{1 + \sin x} \]

### Step 8: Evaluate the limit

Substituting \( \csc x = \frac{1}{\sin x} \):

\[ \lim_{x \to 0} \frac{1 - \cos x}{\cos x (1 + \sin x)} \]

As \( x \to 0 \):

- \( 1 - \cos(0) = 0 \)
- \( \cos(0) = 1 \)
- \( 1 + \sin(0) = 1 \)

Thus, we have:

\[ \lim_{x \to 0} \frac{0}{1} = 0 \]

### Step 9: Final calculation

Now, substituting back into our limit expression:

\[ \lim_{x \to 0} f(x) = e^0 = 1 \]

### Conclusion

To make \( f(x) \) continuous at \( x = 0 \), we need:

\[ f(0) = 1 \]

Thus, the value of \( f(0) \) that makes \( f(x) \) continuous at \( x = 0 \) is:

\[ \boxed{1} \]