Usinfg L' Hospital's rule, evaluate :
`lim_(xrarr0)(xe^(x)-"log"(1+x))/x^(2)`

Answer

To evaluate the limit \( \lim_{x \to 0} \frac{xe^x - \log(1+x)}{x^2} \) using L'Hôpital's rule, we will follow these steps:

### Step 1: Check the form of the limit
First, we need to check if the limit is in an indeterminate form (0/0 or ∞/∞).

- As \( x \to 0 \):
- The term \( xe^x \) approaches \( 0 \cdot e^0 = 0 \).
- The term \( \log(1+x) \) approaches \( \log(1) = 0 \).
- The denominator \( x^2 \) approaches \( 0 \).

Since both the numerator and denominator approach 0, we have the indeterminate form \( \frac{0}{0} \), which allows us to apply L'Hôpital's rule.

### Step 2: Differentiate the numerator and denominator
We will differentiate the numerator and the denominator separately.

- The derivative of the numerator \( xe^x - \log(1+x) \):
- Using the product rule for \( xe^x \):
\[ \frac{d}{dx}(xe^x) = e^x + xe^x \]
- The derivative of \( \log(1+x) \) is:
\[ \frac{d}{dx}(\log(1+x)) = \frac{1}{1+x} \]
- Therefore, the derivative of the numerator is:
\[ e^x + xe^x - \frac{1}{1+x} \]

- The derivative of the denominator \( x^2 \) is:
\[ \frac{d}{dx}(x^2) = 2x \]

### Step 3: Apply L'Hôpital's Rule
Now we can apply L'Hôpital's rule:
\[ \lim_{x \to 0} \frac{xe^x - \log(1+x)}{x^2} = \lim_{x \to 0} \frac{e^x + xe^x - \frac{1}{1+x}}{2x} \]

### Step 4: Evaluate the new limit
Next, we substitute \( x = 0 \) into the new limit:
- The numerator becomes:
\[ e^0 + 0 \cdot e^0 - \frac{1}{1+0} = 1 + 0 - 1 = 0 \]
- The denominator becomes:
\[ 2 \cdot 0 = 0 \]

Since we still have the form \( \frac{0}{0} \), we apply L'Hôpital's rule again.

### Step 5: Differentiate again
Now we differentiate the new numerator and denominator again:
- The derivative of the numerator \( e^x + xe^x - \frac{1}{1+x} \):
- The derivative of \( e^x \) is \( e^x \).
- The derivative of \( xe^x \) is \( e^x + xe^x \).
- The derivative of \( -\frac{1}{1+x} \) is:
\[ \frac{1}{(1+x)^2} \]
- So, the new numerator is:
\[ e^x + (e^x + xe^x) + \frac{1}{(1+x)^2} = 2e^x + xe^x + \frac{1}{(1+x)^2} \]

- The derivative of the denominator \( 2x \) is:
\[ 2 \]

### Step 6: Apply L'Hôpital's Rule again
Now we apply L'Hôpital's rule again:
\[ \lim_{x \to 0} \frac{2e^x + xe^x + \frac{1}{(1+x)^2}}{2} \]

### Step 7: Evaluate the limit
Substituting \( x = 0 \) gives:
- The numerator becomes:
\[ 2e^0 + 0 \cdot e^0 + \frac{1}{(1+0)^2} = 2 + 0 + 1 = 3 \]
- The denominator is \( 2 \).

Thus, the limit evaluates to:
\[ \frac{3}{2} \]

### Final Answer
The final answer is:
\[ \lim_{x \to 0} \frac{xe^x - \log(1+x)}{x^2} = \frac{3}{2} \]