Question

If $f\left(x\right)$ 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

Given that f(x) is continuous at $x=0$

$\Rightarrow f\left(0\right)={lim}_{x\to 0}f\left(x\right)=2$

$\Rightarrow li{m}_{x\to 0}\frac{8^x-2^x}{k^x-1}=2$

This is in the form of $\frac{0}{0}$, using L'Hospital's reule, we get

$\Rightarrow {lim}_{x\to 0}\frac{8^x\log 8-2^x\log 2}{k^x\log k}=2$

$\Rightarrow \frac{\log 8-\log 2}{\log k}=2$

$\Rightarrow \frac{3\log 2-\log 2}{\log k}=2$

$\Rightarrow \frac{2\log 2}{\log k}=2$

Therefore, $\Rightarrow k=2$