Evaluate :
`lim_(x to 0) (e^(x)-1)/x `

Answer

To evaluate the limit \( \lim_{x \to 0} \frac{e^x - 1}{x} \), we can use the series expansion of \( e^x \).

### Step-by-Step Solution:

1. **Recall the series expansion of \( e^x \)**:
The series expansion for \( e^x \) is given by:
\[ e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots \]

2. **Substitute the series into the limit**:
We can substitute this series into our limit:
\[ \lim_{x \to 0} \frac{e^x - 1}{x} = \lim_{x \to 0} \frac{\left(1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots\right) - 1}{x} \]

3. **Simplify the expression**:
The \( +1 \) and \( -1 \) cancel each other out:
\[ = \lim_{x \to 0} \frac{\frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots}{x} \]

4. **Factor out \( x \)**:
Dividing each term in the numerator by \( x \):
\[ = \lim_{x \to 0} \left(\frac{x}{x \cdot 1!} + \frac{x^2}{x \cdot 2!} + \frac{x^3}{x \cdot 3!} + \ldots\right) \]
This simplifies to:
\[ = \lim_{x \to 0} \left(\frac{1}{1!} + \frac{x}{2!} + \frac{x^2}{3!} + \ldots\right) \]

5. **Evaluate the limit**:
As \( x \to 0 \), all terms containing \( x \) will approach 0:
\[ = \frac{1}{1!} + 0 + 0 + \ldots = 1 \]

Thus, the final answer is:
\[ \lim_{x \to 0} \frac{e^x - 1}{x} = 1 \]