Using L Hospital rule evaluate `lim_(xrarr0) (e^(x)-1-x)/(x(e^(x)-1))`

Answer

To evaluate the limit

\[ \lim_{x \to 0} \frac{e^x - 1 - x}{x(e^x - 1)}, \]

we can apply L'Hôpital's Rule since both the numerator and denominator approach 0 as \( x \) approaches 0, creating a \( \frac{0}{0} \) indeterminate form.

### Step 1: Differentiate the Numerator and Denominator

1. **Numerator**: \( e^x - 1 - x \)
- The derivative of \( e^x \) is \( e^x \).
- The derivative of \( -1 \) is \( 0 \).
- The derivative of \( -x \) is \( -1 \).
- Therefore, the derivative of the numerator is:
\[ \frac{d}{dx}(e^x - 1 - x) = e^x - 1. \]

2. **Denominator**: \( x(e^x - 1) \)
- We will use the product rule here. Let \( u = x \) and \( v = e^x - 1 \).
- The derivative of \( u \) is \( 1 \).
- The derivative of \( v \) is \( e^x \).
- By the product rule:
\[ \frac{d}{dx}(x(e^x - 1)) = u'v + uv' = 1(e^x - 1) + x(e^x) = e^x - 1 + xe^x. \]

### Step 2: Rewrite the Limit

Now we can rewrite the limit using the derivatives we found:

\[ \lim_{x \to 0} \frac{e^x - 1}{e^x - 1 + xe^x}. \]

### Step 3: Evaluate the Limit Again

Now, we substitute \( x = 0 \):

- The numerator becomes \( e^0 - 1 = 1 - 1 = 0 \).
- The denominator becomes \( e^0 - 1 + 0 \cdot e^0 = 1 - 1 + 0 = 0 \).

This is again a \( \frac{0}{0} \) form, so we apply L'Hôpital's Rule again.

### Step 4: Differentiate Again

1. **Numerator**: \( e^x - 1 \)
- The derivative is \( e^x \).

2. **Denominator**: \( e^x - 1 + xe^x \)
- The derivative of \( e^x - 1 \) is \( e^x \).
- The derivative of \( xe^x \) (using the product rule) is:
\[ e^x + xe^x. \]
- Therefore, the derivative of the denominator is:
\[ e^x + (e^x + xe^x) = 2e^x + xe^x. \]

### Step 5: Rewrite the Limit Again

Now we have:

\[ \lim_{x \to 0} \frac{e^x}{2e^x + xe^x}. \]

### Step 6: Evaluate the Final Limit

Substituting \( x = 0 \):

- The numerator becomes \( e^0 = 1 \).
- The denominator becomes \( 2e^0 + 0 \cdot e^0 = 2 + 0 = 2 \).

Thus, the limit evaluates to:

\[ \frac{1}{2}. \]

### Final Answer

The limit is

\[ \frac{1}{2}. \]