Let f: R → R be defined as f (x) = \ cases λ| x ^ 2 - 5x + 6| μ(5x...

If $f$\left(x\right)$is continuous at$x = 2$, then$\lim_{x \to {2}^{+}}f\left(x\right) = \lim_{x \to {2}^{-}}f\left(x\right) = \lim_{x \to 2}f\left(x\right)$</p><p>Here,</p><p>$\lim_{x \to {2}^{+}}f\left(x\right) = \lim_{x \to {2}^{+}}e\dfrac{\tan ( x - 2 )}{x - 2} = {e}^{1} = e$</p><p>$\lim_{x \to {2}^{-}}f\left(x\right) = \lim_{x \to {2}^{-}}\dfrac{\lambda\left|{x}^{2} - 5x + 6\right|}{\mu\left(5x - {x}^{2} - 6\right)} = \lim_{x \to {2}^{-}}\dfrac{- \lambda ( x - 2 ) ( x - 3 )}{\mu ( x - 2 ) ( x - 3 )} = - \dfrac{\lambda}{\mu}</p><p>\left(∵ \left|{x}^{2} - 5x + 6\right| = \left\{$\begin{array}{ll}\left(x - 2\right)\left(x - 3\right)\text{if x 3} \\ - \left(x - 2\right)\left(x - 3\right)\text{if 2 < x < 3}\end{array}$\right\right)

For continuity $\mu = e = - \dfrac{\lambda}{\mu} ⇒ \mu = e, \lambda = - {e}^{2}$

$\lambda + \mu = e ( - e + 1 )$