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Category: Limits

lim-x-ln-ax-b-ln-bx-a-

Question Number 1368 by prakash jain last updated on 25/Jul/15 limxln(ax+b)ln(bx+a)=? Commented by 123456 last updated on 25/Jul/15 f(x)=ln(ax+b)ln(bx+a)$$\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} \