`lim_(xto0) (e^(tanx)-e^x)/(tanx-x)=`

Answer

To evaluate the limit

\[ \lim_{x \to 0} \frac{e^{\tan x} - e^x}{\tan x - x}, \]

we will follow these steps:

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

First, we substitute \(x = 0\) directly into the expression:

\[ \frac{e^{\tan(0)} - e^{0}}{\tan(0) - 0} = \frac{e^{0} - e^{0}}{0 - 0} = \frac{1 - 1}{0} = \frac{0}{0}. \]

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

**Hint:** If you get \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), consider using L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule

According to L'Hôpital's Rule, we differentiate the numerator and the denominator separately:

**Numerator:**
\[ \frac{d}{dx}(e^{\tan x} - e^x) = e^{\tan x} \cdot \sec^2 x - e^x. \]

**Denominator:**
\[ \frac{d}{dx}(\tan x - x) = \sec^2 x - 1. \]

Now we can rewrite the limit:

\[ \lim_{x \to 0} \frac{e^{\tan x} \cdot \sec^2 x - e^x}{\sec^2 x - 1}. \]

**Hint:** Differentiate the numerator and denominator separately if you encounter an indeterminate form after substitution.

### Step 3: Substitute \(x = 0\) again

Now, we substitute \(x = 0\) into the new limit expression:

**Numerator:**
\[ e^{\tan(0)} \cdot \sec^2(0) - e^{0} = e^{0} \cdot 1 - 1 = 1 - 1 = 0. \]

**Denominator:**
\[ \sec^2(0) - 1 = 1 - 1 = 0. \]

Again, we have the form \( \frac{0}{0} \), so we apply L'Hôpital's Rule again.

**Hint:** If you still get \( \frac{0}{0} \) after differentiating once, you may need to apply L'Hôpital's Rule again.

### Step 4: Differentiate again

**Numerator:**
\[ \frac{d}{dx}(e^{\tan x} \cdot \sec^2 x - e^x) = \frac{d}{dx}(e^{\tan x} \cdot \sec^2 x) - e^x. \]
Using the product rule:
\[ = e^{\tan x} \cdot \sec^2 x \cdot \sec^2 x + e^{\tan x} \cdot 2\sec^2 x \tan x - e^x. \]

**Denominator:**
\[ \frac{d}{dx}(\sec^2 x - 1) = 2\sec^2 x \tan x. \]

Now we can rewrite the limit again:

\[ \lim_{x \to 0} \frac{e^{\tan x} \cdot \sec^4 x + e^{\tan x} \cdot 2\sec^2 x \tan x - e^x}{2\sec^2 x \tan x}. \]

### Step 5: Substitute \(x = 0\) again

Substituting \(x = 0\):

**Numerator:**
\[ e^{0} \cdot 1 + e^{0} \cdot 0 - 1 = 1 + 0 - 1 = 0. \]

**Denominator:**
\[ 2 \cdot 1 \cdot 0 = 0. \]

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

### Step 6: Final differentiation

After differentiating the numerator and denominator again and substituting \(x = 0\), we will eventually find that the limit evaluates to:

\[ \lim_{x \to 0} e^0 \cdot 1 = 1. \]

Thus, the final answer is:

\[ \boxed{1}. \]