Question

Let \( f(x) \) be differentiable on \( \mathbb{R} \) and \( f'(m) \ne 0, m \in \mathbb{R} \). If \( \lim\limits_{x \to m} \frac{x f(m) - m f(x)}{x - m} + f'(m) = f(m) \), then \( m = \):

• A. 0
• B. -1
• C. 1
• D. 2

Hint:

When handling limits involving functions and their derivatives, try simplifying using properties of differentiability and algebraic manipulation.

Answer 1

The Correct Option is C

We are given: \[ \lim_{x \to m} \frac{x f(m) - m f(x)}{x - m} + f'(m) = f(m) \] Split the limit: \[ \frac{x f(m) - m f(x)}{x - m} = \frac{x f(m) - m f(m) + m f(m) - m f(x)}{x - m} = \frac{(x - m) f(m)}{x - m} + m \cdot \frac{f(m) - f(x)}{x - m} \] So, the expression becomes: \[ f(m) - m f'(m) \] Now, equating: \[ f(m) - m f'(m) + f'(m) = f(m) \Rightarrow -m f'(m) + f'(m) = 0 \Rightarrow f'(m)(1 - m) = 0 \] Since \( f'(m) \ne 0 \), we must have: \[ 1 - m = 0 \Rightarrow m = 1 \]