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Any ideas on how to calculate $\lim_{x \to 0} \left(\frac{1}{e^x-1}-\frac{1}{x}\right)$ without using l'Hopital's rule?

I tried putting $u = e^x-1$ and $x = \ln(u+1)$, replacing but i dont get much farther than that. I wanted to use the limit $\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x = e$ along to solve it but dont know how.

$$\lim_{x \to 0} \left(\frac{1}{e^x-1}-\frac{1}{x}\right) = \lim_{u \to 0} \left(\frac{1}{u}-\frac{1}{\ln(u+1)}\right) = \lim_{u \to 0} \left(\frac{\ln(u+1)-u}{u\cdot \ln(u+1)}\right) $$

If the $u$ multiplying in the denominator were to be $ \frac{1}{u} $, I could make $\frac{1}{u} \cdot \ln(u+1) = \ln(u+1)^\frac{1}{u}$ and do $t = \frac{1}{u} $ (with $ {u \to 0} $ implying $ {t \to \infty} $) and $ \ln(u+1)^\frac{1}{u} = \ln(\frac{1}{t} + 1)^t $. Then $ \lim_{t \to \infty} \ln(\frac{1}{t} + 1)^t = \ln(\lim_{t \to \infty} \left(\frac{1}{t} + 1)^t\right) = \ln(e) = 1 $. That is as far as I got.

user
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creepshow
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    You need to show your effort for what you've tried to answer the question. This isn't a homework service website. – Adam Rubinson Jun 19 '23 at 21:12
  • Hi, welcome to Math SE. Can you use $e^x-1-x\sim\frac12x^2$? – J.G. Jun 19 '23 at 21:13
  • @J.G. Hi, thanks! i dont think so, it's a calculus 1 course problem. – creepshow Jun 19 '23 at 21:19
  • @creepshow Welcome to stack exchange. If you show what you have done so far, we will help you. As Adam said, this isn't a homework service. We need to know that you have actually put effort into the problem before posting it. – Dylan Levine Jun 19 '23 at 21:28
  • @DylanLevine thanks, i didnt know. i already edited how i thought to solve it – creepshow Jun 19 '23 at 21:32
  • @creepshow Welcome. Try to follow JG's comment above to find the limit $-0.5$ If you know l'hopital's rule,and $e$ it self, you must know too that approximation of $e$. – Piquito Jun 19 '23 at 21:38
  • https://math.stackexchange.com/questions/3348754/lim-x-to-0-left-frac1x-frac1ex-1-right?noredirect=1 – Sine of the Time Jun 19 '23 at 21:44
  • @Piquito I think that is a very general assumption to make. I just finished AP Calc AB and I know about L'Hopital's rule and about $e$ but the approximation is unfortunately not in the curriculum. – Dylan Levine Jun 19 '23 at 21:46
  • It has a "duplicate" closing vote. It's strictly true, but, what if the answers for the older question are not satisfactory? (I think they are not and, anyway, this is reasonably opinable). The "lack of..." closing votes seem to be premature as the OP edited the question in a fashion that can be considered acceptable. – Rafa Budría Jun 19 '23 at 22:22

2 Answers2

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The substitution you tried can't help so much, instead we can try with

$$\frac{1}{e^x-1}-\frac{1}{x} =\frac{x-e^x+1}{x(e^x-1)}=-\frac{x}{e^x-1}\frac{e^x-x-1}{x^2}$$

which reduces to the standard limit $\lim_{x\to 0}\frac{e^x-1}x$ and to

$$\lim_{x\to 0}\frac{e^x-x-1}{x^2}$$

which is discussed here

Note that the way you were looking for by substitution leads to a similar solution indeed from here

$$\ldots=\frac{\ln(u+1)-u}{u\cdot \ln(u+1)}=\frac{u}{\ln (1+u)}\frac{\ln(1+u)-u}{u^2}$$

which reduces to the standard limit $\lim_{x\to 0}\frac{\ln(1+x)}x$ and to

$$\lim_{x\to 0}\frac{\ln(1+x)-x}{x^2}$$

which is discussed in the same reference.

user
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  • You beat me to it :) – Dylan Levine Jun 19 '23 at 21:37
  • Please post your own answer! I can delete that one.Bye – user Jun 19 '23 at 21:38
  • No, please keep it up. I was just joking and your answer is much better than mine would have been – Dylan Levine Jun 19 '23 at 21:42
  • how do i get the first step? because i cant make the denominator. i had e^x-1 but in the denominator there it is e^x+1 – creepshow Jun 19 '23 at 21:44
  • maybe i'm not seeing it but i think the first two expressions on the first equal are not equal – creepshow Jun 19 '23 at 21:45
  • @creepshow It was a typo, now fixed. By substitution you were going in a similar solution.I add something on that. – user Jun 19 '23 at 21:45
  • thanks, although now i'm having lots of problems understanding the fist response of the reference. the second response there is not going to be useful because in the course we haven't seen binomial theorem. – creepshow Jun 19 '23 at 21:57
  • @creepshow Then you can refer to the method by substitution,using the limit with $\log$. Is that way clear to you? – user Jun 19 '23 at 21:59
  • @user for instance, in the reference, when its solving L4, how do they make it to the second step? i think when they write "4L4" its because they multiply by 4 the limit but i just dont make sense of the multiplication :( – creepshow Jun 19 '23 at 22:06
  • @creepshow They are plugging $2x$ instead of $x$ nad the $4$ comes out form the denominator. – user Jun 19 '23 at 22:10
  • @creepshow He made the change of variable $x\to 2x$, then got rid of the 4 in the denominator multiplicating the limit by 4. – Rafa Budría Jun 19 '23 at 22:10
  • oh! i didnt see it. thanks you so much! – creepshow Jun 19 '23 at 22:16
  • then i'm already lost from step 2 to step 3 – creepshow Jun 19 '23 at 22:24
  • @creepshow He calculated $3L_4=4L_4-L_4$ from the previous values (then, again a variable change, etc.) – Rafa Budría Jun 19 '23 at 22:28
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Asymptotics ... As $x \to 0$, \begin{align} e^x &= 1+x+\frac{1}{2}x^2 + o(x^2) \\ e^x-1 &= x+\frac{1}{2}x^2+o(x^2) = x\left(1+\frac{1}{2}x + o(x)\right) \\ \frac{1}{e^x-1} &= \frac{1}{x}(1-\frac{1}{2}x + o(x)) = \frac{1}{x}-\frac{1}{2} + o(1) \\ \frac{1}{e^x-1} - \frac{1}{x} &= -\frac{1}{2} + o(1) \\ \lim_{x\to 0}\left(\frac{1}{e^x-1} - \frac{1}{x}\right) &= -\frac{1}{2} \end{align}

GEdgar
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  • The OP mentiones in the comments that Taylor polynomials cannot be used. Furthermore, this solution has already been given in the duplicate questions. – Gary Jun 20 '23 at 00:27