Oblicz granicę ciągu: \(\lim_{x \to 0}\frac{4^x-8^x}{7^x-5^x}\)
\(\frac{\ln \dfrac{1}{2}}{\ln \dfrac{7}{5}}\)
\[ \begin{split} &\lim_{x \to 0}\frac{4^x-8^x}{7^x-5^x}=\\[6pt] &=\lim_{x \to 0}\frac{4^x+1-1-8^x}{7^x+1-1-5^x}=\\[6pt] &=\lim_{x \to 0}\dfrac{4^x-1-(8^x-1):x}{7^x-1-(5^x-1):x}=\\[6pt] &=\lim_{x \to 0}\dfrac{\dfrac{4^x-1}{x}-\dfrac{8^x-1}{x}}{\dfrac{7^x-1}{x}-\dfrac{5^x-1}{x}}=\\[6pt] &=\lim_{x \to 0}\frac{\ln 4-\ln 8}{\ln 7-\ln 5}=\\[6pt] &=\frac{\ln \dfrac{1}{2}}{\ln \dfrac{7}{5}} \end{split} \]