\( \lim _{x \rightarrow a} \frac{x\tan a-a\tan x}{x-a} \)

To evaluate the limit \[ L = \lim_{x \rightarrow a} \frac{x \tan a - a \tan x}{x - a}, \] we can start by substituting \( x = a \) directly into the expression. This gives us: \[ L = \frac{a \tan a - a \tan a}{a - a} = \frac{0}{0}, \] which is an indeterminate form. To resolve this, we can apply L'Hôpital's Rule, which is applicable for limits of the form \( \frac{0}{0} \). L'Hôpital's Rule states that if we have a limit of the form \( \frac{0}{0} \), we can take the derivative of the numerator and the derivative of the denominator: 1. Differentiate the numerator: \( f(x) = x \tan a - a \tan x \). - The derivative is: \[ f'(x) = \tan a - a \sec^2 x. \] 2. Differentiate the denominator: \( g(x) = x - a \). - The derivative is: \[ g'(x) = 1. \] Now we can apply L'Hôpital's Rule: \[ L = \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)} = \lim_{x \rightarrow a} \left( \tan a - a \sec^2 x \right). \] Substituting \( x = a \) into the limit gives: \[ L = \tan a - a \sec^2 a. \] Thus, the final result is: \[ \boxed{\tan a - a \sec^2 a}. \]