Examine the continuity of the function $f(x) = \begin{cases} \frac{|x - 4|}{2(x - 4)}, & \text{if } x \neq 4 \\ 0, & \text{if } x = 4 \end{cases}$ at $x = 4$. |
Continuous at $x = 4$ because $f(4) = 0$. Discontinuous at $x = 4$ because the limit exists but is not equal to $0$. Discontinuous at $x = 4$ because $\text{LHL} = -0.5$ and $\text{RHL} = 0.5$. Continuous at $x = 4$ because the function is defined for all real numbers. |
Discontinuous at $x = 4$ because $\text{LHL} = -0.5$ and $\text{RHL} = 0.5$. |
The correct answer is Option (3) → Discontinuous at $x = 4$ because $\text{LHL} = -0.5$ and $\text{RHL} = 0.5$. ## We have, $f(x) = \begin{cases} \frac{|x - 4|}{2(x - 4)}, & \text{if } x \neq 4 \\ 0, & \text{if } x = 4 \end{cases} \text{ at } x = 4$ At $x = 4$, $\text{LHL} = \lim\limits_{x \to 4^-} \frac{|x - 4|}{2(x - 4)}$ Put $x = 4 - h$, $= \lim\limits_{h \to 0} \frac{|4 - h - 4|}{2[(4 - h) - 4]} = \lim\limits_{h \to 0} \frac{|-h|}{(8 - 2h - 8)}$ $= \lim\limits_{h \to 0} \frac{h}{-2h} = -\frac{1}{2} \quad \text{and } f(4) = 0 \neq \text{LHL [given]}$ So, $f(x)$ is discontinuous at $x = 4$. |