If a is the point of discontinuity of the fun | vedclass

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(A) The function $f(x)$ is defined as: $f(x) = \begin{cases} \cos 2 x, & -\infty < x < 0 \ e^{3 x}, & 0 \leq x < 3 \ x^2-4 x+3, & 3 \leq x \leq 6 \

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(A) The function $f(x)$ is defined as: $f(x) = \begin{cases} \cos 2 x, & -\infty 6 \end{cases}$ To find the point of discontinuity $a$, we check the points where the definition changes, specifically $x=0, 3, 6$. At $x=3$: $\lim _{x ightarrow 3^{-}} f(x) = \lim _{x ightarrow 3^{-}} e^{3 x} = e^9$ $\lim _{x ightarrow 3^{+}} f(x) = \lim _{x ightarrow 3^{+}} (x^2-4 x+3) = 3^2 - 4(3) + 3 = 9 - 12 + 3 = 0$ Since $\lim _{x ightarrow 3^{-}} f(x) 
eq \lim _{x ightarrow 3^{+}} f(x)$, the function is discontinuous at $x=3$. Thus, $a=3$. Now, we evaluate the limit: $\lim _{x ightarrow 3} \frac{x^2-9}{x^3-5 x^2+9 x-9} = \lim _{x ightarrow 3} \frac{(x-3)(x+3)}{(x-3)(x^2-2 x+3)}$ $= \lim _{x ightarrow 3} \frac{x+3}{x^2-2 x+3} = \frac{3+3}{3^2-2(3)+3} = \frac{6}{9-6+3} = \frac{6}{6} = 1$

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