Evaluate: `lim_(xto0)(log(e^(x)-x))/(1-cosx)`

Options

A. 0
B. 1
C. 2
D. 3

Answer

To evaluate the limit

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

we begin by substituting \( x = 0 \) into the expression:

1. **Substitution**:
\[ \text{Numerator: } \log(e^0 - 0) = \log(1) = 0, \]
\[ \text{Denominator: } 1 - \cos(0) = 1 - 1 = 0. \]

This results in the indeterminate form \( \frac{0}{0} \).

2. **Applying L'Hospital's Rule**: Since we have the indeterminate form \( \frac{0}{0} \), we can apply L'Hospital's Rule, which states that we can take the derivative of the numerator and the denominator separately.

- **Differentiate the numerator**:
\[ \frac{d}{dx}[\log(e^x - x)] = \frac{1}{e^x - x} \cdot \frac{d}{dx}(e^x - x) = \frac{1}{e^x - x}(e^x - 1). \]

- **Differentiate the denominator**:
\[ \frac{d}{dx}[1 - \cos x] = \sin x. \]

Therefore, we can rewrite the limit as:
\[ \lim_{x \to 0} \frac{\frac{e^x - 1}{e^x - x}}{\sin x}. \]

3. **Substituting Again**: Now, we substitute \( x = 0 \) again:
- The numerator becomes:
\[ e^0 - 1 = 0, \]
and the denominator becomes:
\[ \sin(0) = 0. \]
This again results in the indeterminate form \( \frac{0}{0} \).

4. **Applying L'Hospital's Rule Again**: We apply L'Hospital's Rule a second time:
- **Differentiate the numerator**:
\[ \frac{d}{dx}[e^x - 1] = e^x. \]
- **Differentiate the denominator**:
\[ \frac{d}{dx}[\sin x] = \cos x. \]

Now, we rewrite the limit:
\[ \lim_{x \to 0} \frac{e^x}{\cos x}. \]

5. **Final Substitution**: Substitute \( x = 0 \):
\[ \frac{e^0}{\cos(0)} = \frac{1}{1} = 1. \]

Thus, the value of the limit is

\[ \boxed{1}. \]