Question Number 1368 by prakash jain last updated on 25/Jul/15
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\mathrm{ln}\left({ax}+{b}\right)}{\mathrm{ln}\:\left({bx}+{a}\right)}\:=? \\ $$
Commented by 123456 last updated on 25/Jul/15
$${f}\left({x}\right)=\frac{\mathrm{ln}\:\left({ax}+{b}\right)}{\mathrm{ln}\:\left({bx}+{a}\right)} \\ $$$$\frac{\frac{{d}}{{dx}}\left[\mathrm{ln}\:\left({ax}+{b}\right)\right]}{\frac{{d}}{{dx}}\left[\mathrm{ln}\:\left({bx}+{a}\right)\right]}=\frac{\frac{{a}}{{ax}+{b}}}{\frac{{b}}{{bx}+{a}}}=\frac{{a}\left({bx}+{a}\right)}{{b}\left({ax}+{b}\right)}\rightarrow\frac{{ab}}{{ba}}=\mathrm{1} \\ $$
Answered by 112358 last updated on 25/Jul/15