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(x)=((ln (ax+b))/(ln (bx+a))) (((d/dx)[ln (ax+b)])/((d/dx)[ln (bx+a)]))=((a/(ax+b))/(b/(bx+a)))=((a(bx+a))/(b(ax+b)))→((ab)/(ba))=1](Q1370.png)
$${f}\left({x}\right)=\frac{\mathrm{ln}\:\left({ax}+{b}\right)}{\mathrm{ln}\:\left({bx}+{a}\right)} \\ $$$$\frac{\frac{{d}}{{dx}}\left[\mathrm{ln}\:\left({ax}+{b}\right)\right]}{\frac{{d}}{{dx}}\left[\mathrm{ln}\:\left({bx}+{a}\right)\right]}=\frac{\frac{{a}}{{ax}+{b}}}{\frac{{b}}{{bx}+{a}}}=\frac{{a}\left({bx}+{a}\right)}{{b}\left({ax}+{b}\right)}\rightarrow\frac{{ab}}{{ba}}=\mathrm{1} \\ $$
Answered by 112358 last updated on 25/Jul/15
$${L}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{{ln}\left({ax}+{b}\right)}{{ln}\left({bx}+{a}\right)}=\frac{\infty}{\infty}\:{which}\:{is}\:{indeterminate}. \\ $$$${Perhaps}\:{L}'{H}\hat {{o}ptial}'{s}\:{rule}\:{can}\:{be}\:{used}. \\ $$$$ \\ $$$$ \\ $$
Commented by 123456 last updated on 26/Jul/15
$${a}>\mathrm{0},{b}>\mathrm{0} \\ $$