Evaluate `lim_(xto0) (e^(x)-1-x)/(x^(2)).`

Explanation

To evaluate the limit

\[ \lim_{x \to 0} \frac{e^x - 1 - x}{x^2}, \]

we will follow these steps:

### Step 1: Direct Substitution
First, we will substitute \( x = 0 \) directly into the expression.

\[ e^0 - 1 - 0 = 1 - 1 - 0 = 0, \]
and the denominator becomes \( 0^2 = 0 \).

Since we have the indeterminate form \( \frac{0}{0} \), we need to use another method.

**Hint:** When you get \( \frac{0}{0} \), consider using Taylor series expansion or L'Hôpital's Rule.

### Step 2: Taylor Series Expansion
We can use the Taylor series expansion for \( e^x \) around \( x = 0 \):

\[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \]

Subtracting \( 1 \) and \( x \) from both sides gives:

\[ e^x - 1 - x = \frac{x^2}{2} + \frac{x^3}{6} + \cdots \]

**Hint:** Use the series expansion to simplify the numerator.

### Step 3: Substitute the Expansion into the Limit
Now, substitute the expansion into the limit:

\[ \lim_{x \to 0} \frac{\frac{x^2}{2} + \frac{x^3}{6} + \cdots}{x^2} \]

**Hint:** Factor out \( x^2 \) from the numerator.

### Step 4: Simplify the Expression
We can factor \( x^2 \) out of the numerator:

\[ \lim_{x \to 0} \frac{x^2 \left(\frac{1}{2} + \frac{x}{6} + \cdots\right)}{x^2} \]

This simplifies to:

\[ \lim_{x \to 0} \left(\frac{1}{2} + \frac{x}{6} + \cdots\right) \]

**Hint:** Evaluate the limit by substituting \( x = 0 \) into the simplified expression.

### Step 5: Evaluate the Limit
Now, substituting \( x = 0 \):

\[ \frac{1}{2} + 0 + \cdots = \frac{1}{2} \]

Thus, the limit is:

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

### Final Answer
\[ \boxed{\frac{1}{2}}. \]