Question

Examine the continuity of the function:
$f\left(x\right)=\frac{\log 100+\log (0.01+x)}{3x}$, for $x\ne 0=\frac{100}{3}$
for $x=0$; at $x=0$.

Answer 1

For $x\ne 0,f\left(x\right)=\frac{\log 100+\log (0.01+x)}{3x}=\frac{\log (1+100x)}{3x}$

For continuity at $x=0$, $f\left(0\right)=f\left(0^{-}\right)=f\left(0^{+}\right)$

$\therefore \underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}\frac{\log (1+100x)}{3x}=\underset{x\to 0^{-}}{lim}\frac{1}{(1+100x)(-3)}(-100)=\frac{100}{3}$

(Using L'Hospital's rule)

$\therefore \underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}\frac{\log (1+100x)}{3x}=\underset{x\to 0^{+}}{lim}\frac{1}{(1+100x)3}\left(100\right)=\frac{100}{3}$

(Using L'Hospital's rule)

As $f\left(0\right)=f\left(0^{-}\right)=f\left(0^{+}\right)=\frac{100}{3}$, the function is continuous at $x=0$.