$\begingroup$I'm aware that with L'Hopital's rule, we're dealing with indeterminate forms of $ \lim_{x\to a} \frac{f(x)}{g(x)} $, and that includes re-writing $\lim_{x\to a} \infty - \infty$ into that format. However, I'm not sure how to proceed with the following:
$$ \lim_{x\to \infty} {\ln3x}-{\ln(x+1)} $$
Should I take the derivative first and then simplify for L'Hopital's rule? I'm not sure where to go.
$\endgroup$3 Answers
$\begingroup$$$ \lim_{x\to\infty} {\ln3x}-{\ln(x+1)}= \lim_{x\to\infty} \ln\frac{3x}{x+1} $$ $$=\lim_{x\to\infty} \ln\frac{3}{1+1/x}=\ln3$$
$\endgroup$$\begingroup$Hint: $\ln a-\ln b=\ln(\frac{a}{b})$
$\endgroup$$\begingroup$$\ln(a)-\ln(b)=\ln\frac{a}{b}$
$\endgroup$
