What to do when L'Hopital's rule does not work

Question

I am looking to calculate the value of the function

$\frac{(ax+1)^{-(1+\frac{1}{ax})}\ln(ax+1)}{x}$

when $x \rightarrow 0$, and $a$ is a positive constant. Repeated application of L'Hopital rule keeps giving an indeterminate form. Any suggestion is highly appreciated.

Kavi Rama Murthy · Accepted Answer · 2019-08-20T05:32:55.180

3

$\lim \frac {\ln(ax+1)} x=a$ by L'Hopital's Rule and $\lim (ax+1)^{-(1+\frac 1 {ax})}=e^{-1}$.

edited Aug 20 '19 at 05:32

answered Aug 20 '19 at 00:38

Kavi Rama Murthy

311,013

1

$(1+ax)^{-(1+(1/ax))}=(1+ax)^{-1}(1+ax)^{-1/ax}\to 1\cdot e^{-1}$ so that the desired result is $a/e$. – Paramanand Singh Aug 20 '19 at 02:28
Thank you for the correction, to get $e^{-1}$ from $(1+ax)^{-1/ax}$, is there an identity that you have used? – King008 Aug 20 '19 at 02:39
Never mind the previous comment, it has to be Euler's identity. Appreciate the help. – King008 Aug 20 '19 at 02:56
@ParamanandSingh Thank you for your comment. I have edited the answer. – Kavi Rama Murthy Aug 20 '19 at 05:33

score 0 · Answer 2 · answered Aug 20 '19 at 02:59

You already received good and simple answers for the limit.

If you want more than just the limit, could could use composition of Taylor series. $$A=(1+a x)^{-(1+\frac{1}{a x})}\implies \log(A)=-(1+\frac{1}{a x})\log(1+ax)$$ $$\log(1+ax)=a x-\frac{a^2 x^2}{2}+\frac{a^3 x^3}{3}+O\left(x^4\right)$$ $$\log(A)=-1-\frac{a x}{2}+\frac{a^2 x^2}{6}+O\left(x^3\right)$$ $$A=e^{\log(A)}=\frac{1}{e}-\frac{a x}{2 e}+\frac{7 a^2 x^2}{24 e}+O\left(x^3\right)$$ $$A\log(1+ax)=\frac{a x}{e}-\frac{a^2 x^2}{e}+\frac{7 a^3 x^3}{8 e}+O\left(x^4\right)$$ $$\frac{A\log(1+ax)} x=\frac{a}{e}-\frac{a^2 x}{e}+\frac{7 a^3 x^2}{8 e}+O\left(x^3\right)=\frac{a}{e}-\frac{a^2 x}{e}+O\left(x^2\right)$$ which shows the limit and also how it is approached.

What to do when L'Hopital's rule does not work

2 Answers2