Question

Examine the continuity of the function
$f\left(x\right)=\{\begin{array}{cc}\frac{\log x-\log 7}{x-7}& \text{for}x\ne 7\\ 7,& \text{for}x=7\end{array}$,
at $x=7$.

Answer 1

$f\left(x\right)=\{\begin{array}{c}\frac{\log x-\log 7}{x-7},x\ne 7\\ 7,x=7\end{array}$

Checking continuity at $x=7$

$\Rightarrow \underset{x\to 7^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}\left(\frac{\log (7-h)-\log 7}{7-h-7}\right)=\underset{h\to 0}{lim}\frac{\frac{-1}{(7-h)}+0}{(-1)}=\frac{1}{7}$ ...... $\left(1\right)$ (By L' Hospital Rule)

$\Rightarrow \underset{x\to 7^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}\frac{\log (7+h)-\log 7}{7+h-7}=\underset{h\to 0}{lim}\frac{\frac{1}{7-h}}{1}=\frac{1}{7}\to 2$ (By L' Hospital Rule)

$\Rightarrow f\left(7\right)=7\to 3$

From $\left(1\right),\left(2\right)$ and $\left(3\right)$

$\underset{x\to 7^{-}}{lim}f\left(x\right)=\underset{x\to 7^{+}}{lim}f\left(x\right)=f\left(x\right)\ne f\left(7\right)$

$\Rightarrow f\left(x\right)$ is removable discontinuous function.