Step 1: Understanding the Concept:
The phrase "exactly two roots at \( x = a \)" means that \( x = a \) is a root of multiplicity 2. Algebraically, this implies that \( (x - a)^2 \) is a factor of \( f(x) \), but \( (x - a)^3 \) is not. In calculus terms, this means \( f(a) = 0 \), \( f'(a) = 0 \), but \( f''(a) \neq 0 \). The given limit involves the logarithmic derivative of \( f(x) \).
Step 2: Key Formula or Approach:
1. Let \( f(x) = (x - a)^2 \cdot \phi(x) \), where \( \phi(a) \neq 0 \).
2. Differentiate \( f(x) \) and substitute into the limit expression.
3. Ensure the limit is finite to determine \( \lambda \).
Step 3: Detailed Explanation:
Let's define \( f(x) \) based on its multiplicity at \( a \):
\[ f(x) = (x - a)^2 \phi(x) \quad (\text{where } \phi(a) \neq 0) \]
Differentiating with respect to \( x \) using the product rule:
\[ f'(x) = 2(x - a)\phi(x) + (x - a)^2 \phi'(x) \]
Now, substitute \( f(x) \) and \( f'(x) \) into the expression inside the limit:
\[ \frac{\lambda f'(x)}{f(x)} - \frac{1}{x - a} = \frac{\lambda [2(x - a)\phi(x) + (x - a)^2 \phi'(x)]}{(x - a)^2 \phi(x)} - \frac{1}{x - a} \]
Separate the terms in the numerator of the first fraction:
\[ = \frac{2\lambda (x - a)\phi(x)}{(x - a)^2 \phi(x)} + \frac{\lambda (x - a)^2 \phi'(x)}{(x - a)^2 \phi(x)} - \frac{1}{x - a} \]
Simplify:
\[ = \frac{2\lambda}{x - a} + \frac{\lambda \phi'(x)}{\phi(x)} - \frac{1}{x - a} \]
Combine the terms with the common denominator \( (x - a) \):
\[ = \frac{2\lambda - 1}{x - a} + \frac{\lambda \phi'(x)}{\phi(x)} \]
We are given that the limit as \( x \to a \) is a non-zero finite value \( m \):
\[ \lim_{x \to a} \left( \frac{2\lambda - 1}{x - a} + \frac{\lambda \phi'(x)}{\phi(x)} \right) = m \]
For the limit to be finite, the term \( \frac{2\lambda - 1}{x - a} \) must not diverge. Since \( x - a \to 0 \), the numerator must be zero.
\[ 2\lambda - 1 = 0 \implies \lambda = \frac{1}{2} \]
If \( \lambda = 1/2 \), the first term vanishes, and the limit becomes:
\[ \lim_{x \to a} \frac{1}{2} \cdot \frac{\phi'(x)}{\phi(x)} = \frac{\phi'(a)}{2\phi(a)} = m \]
Since \( m \neq 0 \), this is consistent.
Step 4: Final Answer:
To prevent the limit from blowing up at the double root \( x=a \), the coefficient of the singularity must be nullified. This yields \( \lambda = 1/2 \).