Question

If f(x) is continuous at x=0, where
$f\left(x\right)=\frac{8^x-2^x}{k^x-1},$ for $x\ne 0$
$=2,$ for $x=0$
find k.

Answer 1

$f\left(x\right)=\frac{8^x-2^x}{K^x-1}$, for $x\ne 0$

$=2$, for $x=0$

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

$f\left(0^{+}\right)=f\left(0^{-}\right)=f\left(0\right)$

$f\left(0^{+}\right)=\underset{x\to 0}{lim}\frac{8^x-2^x}{K^x-1}=\underset{x\to 0^{+}}{lim}\frac{2^x[4^x-1]}{K^x-1}$

$=\underset{x\to 0}{lim}\left(\frac{4^x-1}{K^x-1}\right)\left(1\right)[\therefore \underset{x\to 0^{+}}{lim}{2}^x=1]$

$=\underset{x\to 0^{+}}{lim}\left(\frac{(1+x\ln 4-1)}{1+x\ln k-1}\right)(\therefore a^x=\frac{1+x\ln a}{forx\to 0})$

$f\left(0^{+}\right)=\frac{\ln 4}{\ln k}=f\left(0\right)$

$=\frac{2\ln 2}{\ln k}=2$

$\Rightarrow K=2$