Consider a function $f(x)$ which has exactly two roots at $x = a$. If $\lim_{x \to a}\left(\dfrac{\lambda f'(x)}{f(x)} - \dfrac{1}{x-a}\right) = m\,(\neq 0)$, then the value of $\lambda$ is

Since $f(x)$ has exactly two roots at $x = a$, it has a root of multiplicity $2$ at $x = a$. We can write $f(x) = (x-a)^2 g(x)$, where $g(a) \neq 0$ and $g(x)$ is differentiable.

Differentiating with respect to $x$, we get:

$f'(x) = 2(x-a)g(x) + (x-a)^2 g'(x)$

Dividing by $f(x)$, we obtain:

$\dfrac{f'(x)}{f(x)} = \dfrac{2(x-a)g(x) + (x-a)^2 g'(x)}{(x-a)^2 g(x)} = \dfrac{2}{x-a} + \dfrac{g'(x)}{g(x)}$

Substituting this into the given limit expression:

$\lim_{x \to a} \left( \lambda \left( \dfrac{2}{x-a} + \dfrac{g'(x)}{g(x)} \right) - \dfrac{1}{x-a} \right) = m$

$\lim_{x \to a} \left( \dfrac{2\lambda - 1}{x-a} + \lambda \dfrac{g'(x)}{g(x)} \right) = m$

For the limit to exist and be a finite value $m$, the term containing $\dfrac{1}{x-a}$ must vanish because its limit as $x \to a$ is undefined. Therefore, the coefficient of $\dfrac{1}{x-a}$ must be zero:

$2\lambda - 1 = 0$

$\lambda = \dfrac{1}{2}$

With this value of $\lambda$, the limit evaluates to $m = \dfrac{1}{2} \dfrac{g'(a)}{g(a)}$, which can be non-zero as given in the problem.

Answer: $\dfrac{1}{2}$