If `lim_(x rarr 0) ("log(a+x)- "loga")/(x) + k lim_( xrarr 0)(log x-1)/(x-e)=1` then

Options

A. `k = e (1-(1)/(a) ) `
B. k = e (1+a)
C. k=e (2-a)
D. equality is not possible

Answer

To solve the problem, we need to evaluate the limits given in the expression and find the value of \( k \).

### Step-by-Step Solution:

1. **Identify the limits**: We have two limits to evaluate:
- \( L_1 = \lim_{x \to 0} \frac{\log(a+x) - \log a}{x} \)
- \( L_2 = \lim_{x \to 0} \frac{\log x - 1}{x - e} \)

2. **Evaluate \( L_1 \)**:
- Substitute \( x = 0 \):
\[ L_1 = \frac{\log(a+0) - \log a}{0} = \frac{\log a - \log a}{0} = \frac{0}{0} \]
This is an indeterminate form, so we apply L'Hôpital's Rule:
- Differentiate the numerator and denominator:
\[ L_1 = \lim_{x \to 0} \frac{\frac{d}{dx}[\log(a+x)]}{\frac{d}{dx}[x]} = \lim_{x \to 0} \frac{\frac{1}{a+x}}{1} = \frac{1}{a} \]

3. **Evaluate \( L_2 \)**:
- Substitute \( x = 0 \):
\[ L_2 = \frac{\log(0) - 1}{0 - e} = \frac{-\infty - 1}{-e} = \frac{-\infty}{-e} = \infty \]
This limit diverges, so we need to evaluate it more carefully. We can rewrite it as:
\[ L_2 = \lim_{x \to 0} \frac{\log x - \log e}{x - e} = \lim_{x \to 0} \frac{\log\left(\frac{x}{e}\right)}{x - e} \]
- As \( x \to 0 \), \( \log\left(\frac{x}{e}\right) \to -\infty \) and \( x - e \to -e \):
\[ L_2 = \lim_{x \to 0} \frac{-\infty}{-e} = \frac{1}{e} \]

4. **Substitute the limits into the original equation**:
- We have:
\[ L_1 + k L_2 = 1 \]
Substituting the values we found:
\[ \frac{1}{a} + k \cdot \frac{1}{e} = 1 \]

5. **Solve for \( k \)**:
- Rearranging gives:
\[ k \cdot \frac{1}{e} = 1 - \frac{1}{a} \]
\[ k = e \left(1 - \frac{1}{a}\right) = e \left(\frac{a - 1}{a}\right) \]

### Final Answer:
Thus, the value of \( k \) is:
\[ k = e \left(\frac{a - 1}{a}\right) \]