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(C) For $f$ to be continuous at $x = 2$,we must have $\lim_{x \rightarrow 2^{-}} f(x) = \lim_{x \rightarrow 2^{+}} f(x) = f(2) = \mu$.First,calculate

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$e(-e+1)$

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(C) For $f$ to be continuous at $x = 2$,we must have $\lim_{x ightarrow 2^{-}} f(x) = \lim_{x ightarrow 2^{+}} f(x) = f(2) = \mu$.First,calculate the right-hand limit:$\lim_{x ightarrow 2^{+}} f(x) = \lim_{x ightarrow 2^{+}} e^{\frac{	an(x-2)}{x-[x]}}$. Since $x > 2$,$[x] = 2$,so $\lim_{x ightarrow 2^{+}} e^{\frac{	an(x-2)}{x-2}} = e^{1} = e$.Thus,$\mu = e$.Next,calculate the left-hand limit:$\lim_{x ightarrow 2^{-}} f(x) = \lim_{x ightarrow 2^{-}} \frac{\lambda|x^{2}-5x+6|}{\mu(5x-x^{2}-6)}$.Note that $x^{2}-5x+6 = (x-2)(x-3)$. For $x  0$. Thus,$|x^{2}-5x+6| = (x-2)(x-3)$.Also,$5x-x^{2}-6 = -(x^{2}-5x+6) = -(x-2)(x-3)$.So,$\lim_{x ightarrow 2^{-}} f(x) = \frac{\lambda(x-2)(x-3)}{\mu(-(x-2)(x-3))} = -\frac{\lambda}{\mu}$.Equating the limits: $-\frac{\lambda}{\mu} = e \Rightarrow \lambda = -\mu e = -e^{2}$.Therefore,$\lambda + \mu = -e^{2} + e = e(-e+1)$.

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