Question

$\underset{x\to 0}{lim}\frac{5^x-4^x}{4^x-3^x}=\log \frac{a}{4}/\log \frac{b}{3}.$
Find $a+b$

Answer 1

$\underset{x\to 0}{lim}\frac{5^x-4^x}{4^x-3^x}$
$=\underset{x\to 0}{lim}\frac{\frac{5^x-1}{x}-\frac{4^x-1}{x}}{\frac{4^x-1}{x}-\frac{3^x-1}{x}}=\frac{\log 5-\log 4}{\log 4-\log 3}$
$=\frac{\log \frac{5}{4}}{\log \frac{4}{3}}.$
$\Rightarrow a+b=5+4=9$
Note: we can also use L'Hospital's rule directly