Question

$f\left(x\right)=\{\begin{array}{cc}\frac{\log 100+\log (0.01+x)}{3x},& \text{for}x\ne 0\\ \frac{100}{3},& \text{for}x=0\end{array}$.
Check the continuity at $x=0$.

Answer 1

$f\left(x\right)=\{\begin{array}{c}\frac{\log 100+\log (0.01+x)}{3x},x\ne 0\\ \frac{100}{3},x=0\end{array}$

Checking continuity at $x=0$

$\Rightarrow \underset{x\to 0^{+}}{lim}\frac{\log 100+\log (0.01+x)}{3x}=\underset{h\to 0}{lim}\frac{\log 100+\log (0.01+0+h)}{3(0+h)}=\underset{h\to 0}{lim}\frac{\frac{1}{(0.01+h)}}{3\left(1\right)}=\frac{100}{3}$ ... $\left(1\right)$ (By L' Hospital Rule)

$\Rightarrow \underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}\frac{\log 100+\log (0.01+0-h)}{3(0-h)}=\underset{h\to 0}{lim}\frac{0+\frac{-1}{0.01-h}}{3(-1)}=\frac{100}{3}$ .... $\left(2\right)$ (L' Hospital Rule)

$\Rightarrow f\left(0\right)=\frac{100}{3}\to 3$

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

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

$\Rightarrow f\left(x\right)$ is continuous at $x=0$