Examine the continuity of the function f(x) = […] at x = 2.

The correct answer is Option (2) → Continuous at $x = 2$ because $\lim\limits_{x \to 2} f(x) = f(2) = 5$. ##

We have, $f(x) = \begin{cases} \frac{2x^2 - 3x - 2}{x - 2}, & \text{if } x \neq 2 \\ 5, & \text{if } x = 2 \end{cases}$ at $x = 2$

At $x = 2$,

$\text{LHL} = \lim\limits_{x \to 2^-} \frac{2x^2 - 3x - 2}{x - 2}$

Put $x = 2 - h$,

$= \lim\limits_{h \to 0} \frac{2(2 - h)^2 - 3(2 - h) - 2}{(2 - h) - 2}$

$= \lim\limits_{h \to 0} \frac{8 + 2h^2 - 8h - 6 + 3h - 2}{-h}$

$= \lim\limits_{h \to 0} \frac{2h^2 - 5h}{-h} = \lim\limits_{h \to 0} \frac{h(2h - 5)}{-h} = 5$

$\text{RHL} = \lim\limits_{x \to 2^+} \frac{2x^2 - 3x - 2}{x - 2}$

Put $x = 2 + h$,

$= \lim\limits_{h \to 0} \frac{2(2 + h)^2 - 3(2 + h) - 2}{(2 + h) - 2}$

$= \lim\limits_{h \to 0} \frac{8 + 2h^2 + 8h - 6 - 3h - 2}{h}$

$= \lim\limits_{h \to 0} \frac{2h^2 + 5h}{h} = \lim\limits_{h \to 0} \frac{h(2h + 5)}{h} = 5$

And $f(2) = 5$

$∴\text{LHL} = \text{RHL} = f(2)$

So, $f(x)$ is continuous at $x = 2$.