Question

$\underset{x\to 0}{lim}\frac{8^x-4^x-2^x+1^x}{x^2}$ is $\log \left(a\right)\log \left(b\right)$Find $a+b$

Answer 1

$\underset{x\to 0}{lim}\frac{8^x-4^x-2^x+1^x}{x^2}=\underset{x\to 0}{lim}\frac{8^x-4^x-2^x+1}{x^2}$

$=\underset{x\to 0}{lim}\frac{(4^x-1)}{x}.\frac{(2^x-1)}{x}=\log 4.\log 2=2{\left(\log 2\right)}^2$

$\left[\text{Using}\underset{x\to 0}{lim}\frac{{\alpha }^x-1}{x}\left(\frac{0}{0}\right)=\underset{x\to 0}{lim}\frac{{\alpha }^x\log \alpha }{1}=\log \alpha \right]$

$\Rightarrow a+b=4+2=6$