Question

Prove that given function is not continuous

$f\left(x\right)=\begin{array}{cc}\frac{\log x-\log 7}{x-7},& forx\ne 7\\ =7,& forx=7\end{array}\}$ at $x=7$.

Answer 1

A function $7\left(x\right)$ is said to be continuous at a point $x=0$, in its domain if the following three conditions are satisfied:

$\to f\left(a\right)$ exists (i.e., the value of $f\left(a\right)$ is finite)

$\to \underset{x\to a}{lim}f\left(x\right)$ exists (i.e. the right hand limit equal to left hand limit & both are finite).

$\to \underset{x\to a}{lim}f\left(x\right)=f\left(a\right)$

LHL for $x=7$

$f\left(x\right)=\left[\frac{log(7-h)-log7}{7-h-7}\right]$ [$h>0$ and limit h tending to zero]

$=\frac{log\left(\frac{7-h}{7}\right)}{-h}$

$=\frac{log[1+(-\frac{h}{7})]}{7(-\frac{h}{7})}=\frac{1}{7}$ $\left[\underset{x\to 0}{lim}\frac{log(1+x)}{x}=1\right]$

RHL for $x=7$

$f\left(x\right)=\left[\frac{log(7+h)-log7}{7+h-7}\right]$ [$h>0$ & limit h tending to zero]

$=\frac{log\left(\frac{7+h}{h}\right)}{h}$

$=\frac{log[1+\left(\frac{h}{7}\right)]}{7\left(\frac{h}{7}\right)}$

$=\frac{1}{7}$

for $x=7$, $f\left(x\right)$ is given as $7$

Hence, LHL$=$RHL but not equal to $f\left(x\right)$

$\therefore f\left(x\right)$ is conti at $x=7$.