Question

Solve $\underset{x\to 0}{lim}\frac{3^x-4^x}{5^x-6^x}$

Answer 1

Given

$\underset{x\to 0}{lim}\left(\frac{3^x-4^x}{5^x-6^x}\right)$

Apply L hopital's rule

$=\underset{x\to 0}{lim}\left(\frac{\ln 3{\cdot 3}^x-\ln 2\cdot 2^{2x+1}}{\ln 5\cdot 5^x-{\ln 6\cdot 6}^x}\right)$

Put $x=0$

$=\left(\frac{\ln 3{\cdot 3}^0-\ln 2\cdot 2^{20+1}}{\ln 5\cdot 5^0-{\ln 6\cdot 6}^0}\right)$

$=\left(\frac{\ln 3-2\ln 2}{\ln 5-\ln 6}\right)$

$=\left(\frac{2\ln 2-\ln 3}{\ln 6-\ln 5}\right)$