Question

Discuss the continuity of the following function at $x=0$. If the function has a removable discontinuity, redefine the function so as to remove the discontinuity.

$f\left(x\right)=\{\begin{array}{c}\frac{4^x-e^x}{6^x-1}\text{for}x\ne 0\\ \log \left(\frac{2}{3}\right)\text{for}x=0\end{array}$

Answer 1

Here, $f\left(0\right)=\log \left(\frac{2}{3}\right)$
Now, $\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{4^x-e^x}{6^x-1}$

$=\underset{x\to 0}{lim}\frac{(4^x-1)-(e^x-1)}{6^x-1}$
$=\underset{x\to 0}{lim}\left[\frac{\frac{(4^x-1)-(e^x-1)}{x}}{\frac{6^x-1}{x}}\right]$

$=\frac{\underset{x\to 0}{lim}\left(\frac{4^x-1}{x}\right)-\underset{x\to 0}{lim}\left(\frac{e^x-1}{x}\right)}{\underset{x\to 0}{lim}\frac{6^x-1}{x}}$
$=\frac{\log 4-\log e}{\log 6}=\frac{\log \left(\frac{4}{e}\right)}{\log 6}$

$\therefore \underset{x\to 0}{lim}f\left(x\right)\ne f\left(0\right)$
$\therefore f\left(x\right)$ is discontinuous at $x=0$.
Here, $\underset{x\to 0}{lim}f\left(x\right)$ exists, but not equal to

$f\left(0\right)$. Hence, the discontinuity at $x=0$ is removable and it can be removed by redefining the function as follows:

$f\left(x\right)=\{\begin{array}{c}\frac{4^x-e^x}{6^x-1}\text{for}x\ne 0\\ \frac{\log \left(\frac{4}{e}\right)}{\log 6}\text{for}x=0\end{array}$