Why the the following limit is $1$? (This result is given by Maple.) $$\lim\limits_{x\rightarrow0}\,\frac{\ln(Ax+Bx^b)}{\ln(x)},$$ where $A>0,B>0,0<b<1$ are contants.
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Have you tried with the L'Hôpital Rule? – LaloVelasco Sep 17 '16 at 04:11
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The limit is not $1$ unless $b=1$. I'm curious about what Maple is doing. – Martin Argerami Sep 17 '16 at 04:18
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But, when you apply L'Hopilal you arrive to $$ \frac{Ax+Bbx^{b}}{Ax+Bx^{b}}. $$ Since $b<1$, the leading term both in the numerator and denominator is $Ax$. Since is the same in both parts, the limit is 1 – LaloVelasco Sep 17 '16 at 04:19
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Precisely because $b<1$, the leading term is $x^b$. – Martin Argerami Sep 17 '16 at 04:28
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@LaloVelasco, thank you for your answer! But once again applying the L'Hopilal rule to the $\frac{Ax+Bbx^b}{Ax+Bx^b}$, the result will be $b$. Will you please help me to explain this result? – Dave Sep 17 '16 at 04:29
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@Martin Argerami, Thank you for your help! – Dave Sep 17 '16 at 04:29
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I have understood my mistake: $x^{b}$ is "larger" than $x$ as $x\rightarrow 0$. Thanks @MartinArgerami! – LaloVelasco Sep 17 '16 at 04:37
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As $1-b>0$, $$ \frac{\log(Ax+Bx^b)}{\log x}=\frac{\log(x^b(Ax^{1-b}+B))}{\log x} =\frac{b\log x+\log(Ax^{1-b}+B)}{\log x} =b+\frac{\log(Ax^{1-b}+B)}{\log x} $$ and then $$ \lim_{x\to0^+}\frac{\log(Ax+Bx^b)}{\log x}=b. $$
The second term goes to zero because the numerator converges to $\log B$.
Martin Argerami
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