Q103 / 160 · TS EAMCET 2019 (03 May Shift 1) · Mathematics
If ' $a$ ' is the point of discontinuity of the function
$$
f(x)=\left\{\begin{array}{cl}
\cos 2 x, & \text { for }-\infty < x < 0 \\
e^{3 x}, & \text { for } 0 \leq x < 3 \\
x^2-4 x+3, & \text { for } 3 \leq x \leq 6 \\
\frac{\log (15 x-89)}{x-6}, & \text { for } x>6
\end{array}\right.
$$
Then, $\lim _{x \rightarrow a} \frac{x^2-9}{x^3-5 x^2+9 x-9}=$
- Correct answerA1
- B0
- C6
- D3
Solution
We have,
$$
\begin{aligned}
& f(x)=\left\{\begin{array}{rc}
\cos 2 x, & -\infty < x < 0 \\
e^{3 x}, & 0 \leq x < 3 \\
x^2-4 x+3, & 3 \leq x \leq 6 \\
\frac{\log (15 x-89)}{x-6}, & x>6
\end{array}\right. \\
& \because \lim _{x \rightarrow 3^{-}} e^{3 x}=e^9 \text { and } \lim _{x \rightarrow 3^{+}} x^2-4 x+3=9-12+3=0
\end{aligned}
$$
Clearly, $\quad \lim _{x \rightarrow 3^{-}} f(x) \neq \lim _{x \rightarrow 3^{+}} f(x)$
$\therefore f(x)$ is discontinuous at $x=3$.
$$
\therefore a=3 \text {, }
$$
Now, $\lim _{x \rightarrow 3} \frac{x^2-9}{x^3-5 x^2+9 x-9}=\lim _{x \rightarrow 3} \frac{(x-3)(x+3)}{(x-3)\left(x^2-2 x+3\right)}$
$$
=\frac{3+3}{(3)^2-2(3)+3}=\frac{2}{2}=1
$$
Hence, the correct answer is option: 'A'.