Question

Find $k$, if $f\left(x\right)=\frac{\log (1+3x)}{5x}$ for $x\ne 0$
$=k$ for $x=0$
is continuous at $x=0$.

Answer 1

$f\left(x\right)=\{\begin{array}{c}\frac{\log (1+3x)}{5x},forx\ne 0\\ k,forx=0\end{array}$

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

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

$\underset{h\to 0}{lim}\frac{\log (1+3(0+h))}{5(0+h)}=\underset{x\to 0^{-}}{lim}\frac{\log (1+3(0-h))}{5(0-h)}=k$

$\Rightarrow \underset{h\to 0}{lim}\frac{\frac{1}{1+3h}.\left(3\right)}{5}=\underset{h\to 0}{lim}\frac{\frac{1}{1-3h}(-3)}{-5}=k$ (L' Hospital Rule)

$\Rightarrow \frac{3}{5}=\frac{3}{5}=k$

$\Rightarrow k=\frac{3}{5}$