`lim_(x to 0) (sin x)/x` =

Options

A. 0
B. 1
C. 2
D. none

Answer

To solve the limit \( \lim_{x \to 0} \frac{\sin x}{x} \), we can use the Taylor series expansion for \( \sin x \) around \( x = 0 \).

### Step-by-Step Solution:

1. **Write the Taylor Series Expansion for \( \sin x \)**:
The Taylor series expansion of \( \sin x \) around \( x = 0 \) is given by:
\[ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \]

2. **Substitute the Expansion into the Limit**:
We substitute the expansion into the limit:
\[ \lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots}{x} \]

3. **Simplify the Expression**:
We can simplify the fraction:
\[ = \lim_{x \to 0} \left( 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \cdots \right) \]

4. **Evaluate the Limit**:
As \( x \) approaches \( 0 \), all terms involving \( x \) will approach \( 0 \):
\[ = 1 - 0 + 0 - \cdots = 1 \]

5. **Conclusion**:
Therefore, we conclude that:
\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]

### Final Answer:
\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]

---