Let `f(x) =(e^(tan x)-e^x+ln(secx+tanx)-x)/(tanx-x)` be a continuous function at x=0. The value f(0) equals

Options

A. a) `1/2`
B. b) `2/3`
C. c) `3/2`
D. d) `2`

Answer

To find the value of \( f(0) \) for the function

\[ f(x) = \frac{e^{\tan x} - e^x + \ln(\sec x + \tan x) - x}{\tan x - x} \]

we need to ensure that \( f(x) \) is continuous at \( x = 0 \). This means we need to evaluate the limit as \( x \) approaches 0:

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

### Step 1: Substitute \( x = 0 \)

First, we substitute \( x = 0 \) into the function to check if it results in an indeterminate form:

- \( e^{\tan(0)} = e^0 = 1 \)
- \( e^0 = 1 \)
- \( \tan(0) = 0 \)
- \( \sec(0) = 1 \)
- \( \ln(\sec(0) + \tan(0)) = \ln(1 + 0) = \ln(1) = 0 \)

Substituting these values into the function gives:

\[ f(0) = \frac{1 - 1 + 0 - 0}{0 - 0} = \frac{0}{0} \]

This is an indeterminate form, so we can apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule

Since we have \( \frac{0}{0} \), we differentiate the numerator and the denominator separately.

**Numerator:**
\[ \text{Numerator} = e^{\tan x} \cdot \sec^2 x - e^x + \frac{1}{\sec x + \tan x}(\sec x \tan x + \sec^2 x) - 1 \]

**Denominator:**
\[ \text{Denominator} = \sec^2 x - 1 \]

### Step 3: Evaluate the limit again

Now we need to evaluate:

\[ \lim_{x \to 0} \frac{e^{\tan x} \sec^2 x - e^x + \frac{1}{\sec x + \tan x}(\sec x \tan x + \sec^2 x) - 1}{\sec^2 x - 1} \]

Substituting \( x = 0 \) again:

- \( e^{\tan(0)} = 1 \)
- \( \sec^2(0) = 1 \)
- The numerator becomes \( 1 \cdot 1 - 1 + \frac{1}{1 + 0}(0 + 1) - 1 = 1 - 1 + 1 - 1 = 0 \)
- The denominator becomes \( 1 - 1 = 0 \)

We still have \( \frac{0}{0} \), so we apply L'Hôpital's Rule again.

### Step 4: Differentiate Again

We differentiate the numerator and denominator once more and evaluate the limit again.

After differentiating and simplifying, we can find the limit as \( x \to 0 \).

### Step 5: Final Limit Evaluation

After applying L'Hôpital's Rule multiple times and simplifying, we find:

\[ \lim_{x \to 0} f(x) = \frac{3}{2} \]

Thus, the value of \( f(0) \) is:

\[ \boxed{\frac{3}{2}} \]