Question

solve $\underset{x\to 0}{lim}\frac{5^x-4^x}{x}$

Answer 1

${lim}_{x\to 0}\frac{5^x-4^x}{x}={lim}_{x\to 0}\frac{(5^x-1)-(4^x-1)}{x}={lim}_{x\to 0}\frac{5^x-1}{x}-{lim}_{x\to 0}\frac{4^x-1}{x}=\log 5-\log 4=\log \frac{5}{4}$