Answer:
Option A
Explanation:
We have ,
f(x)={cos2xfor−∞<x<0e3xfor0≤x<3x2−4x+3,for3≤x≤6log(15x−89)x−6,forx>6
∵ and \lim_{x \rightarrow 3+} x^{2}-4x+3=9-12+3=0
Clearly , \lim_{x \rightarrow 3-}f(x)\neq \lim_{x \rightarrow 3+}f(x)
\therefore f(x) is discontinuous at x=3
\therefore a=3,
Now, \lim_{x \rightarrow 3}\frac{x^{2}-9}{x^{3}-5x^{2}+9x-9}=\lim_{x \rightarrow 3}\frac{(x-3)(x+3)}{(x-3)(x^{2}-2x+3)}
= \frac{3+3}{(3)^{2}-2(3)+3}= \frac{2}{2}=1