## Calculus: Early Transcendentals 8th Edition

$$\lim_{x\to1^+}[\ln(x^7-1)-\ln(x^5-1)]=\ln\frac{7}{5}$$
$$A=\lim_{x\to1^+}[\ln(x^7-1)-\ln(x^5-1)]$$ $$A=\lim_{x\to1^+}\ln\frac{x^7-1}{x^5-1}$$ ($\ln(A-B)=\ln\frac{A}{B}$) $$A=\ln\Big(\lim_{x\to1^+}\frac{x^7-1}{x^5-1}\Big)$$ Since $\lim_{x\to1^+}(x^7-1)=1^7-1=0$ and $\lim_{x\to1^+}[(x^5-1)=1^5-1=0$, we have here an indeterminate form of $\frac{0}{0}$, which we use L'Hospital's Rule to deal with: $$A=\ln\Big[\lim_{x\to1^+}\frac{(x^7-1)'}{(x^5-1)'}\Big]$$ $$A=\ln\Big(\lim_{x\to1^+}\frac{7x^6}{5x^4}\Big)$$ $$A=\ln\frac{7\times1^6}{5\times1^4}$$ $$A=\ln\frac{7}{5}$$