Question

Examine the continuity of $f\left(x\right)$ on its domain where,
$f\left(x\right)=\{\begin{array}{c}\frac{1}{x+1},for2\le x\le 4\\ \frac{x+1}{x-3},4<x\le 6\end{array}$

Answer 1

Since $f\left(x\right)$ is polynomial, it will be continuous throughout the range.

But we need to check its continuity at $x=4$

Left hand limit at $x=4$

$L.H.L.=\underset{x\to 4}{lt}\frac{1}{x+1}=\frac{1}{4+1}=\frac{1}{5}$

Right hand limit at $x=4$

$\begin{array}{c}R.H.L.={lt}_{x\to 4}\frac{x+1}{x-3}=\frac{4+1}{4-3}=5\\ \sin ceL.H.L.\ne R.H.L.atx=4\end{array}$

$f\left(x\right)$ is not continuous at $x=4$

$f\left(x\right)$ is continuous in range

$x\in \left[2,6\right]exceptatx=4$