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Consider the following limit:

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

I want to solve this limit using only elementary tools. This means that definitions of derivatives, integrals, L'Hopital's rule, and series are not allowed. However, you may use the fact that $e^x=\lim_{n\to\infty}\bigg(1+\frac{x}{n}\bigg)^n$

For context, I have been solving typical problems seen in introductory calculus courses, such as $\lim_{x\to0}\frac{x}{3-\sqrt{x+9}}$. The typical way to approach this is by rationalizing the denominator. However, our calculus teacher gave us the problem above, and told us to solve it without using L'Hopital's rule.

The answer our teacher gave us was apparently to graph the equation, but I was not satisfied, and I wanted to try doing it with basic algebraic operations. Apologies if this is too restrictive.

I tried using the substitution $x\to-x$ (which is just $L$) and added the two results:

$$2L=\lim_{x\to 0}\frac{x^{2}}{e^{x}-\left(x+1\right)}+\frac{x^{2}}{e^{-x}+\left(x-1\right)}$$

$$L=\lim_{x\to 0}\frac{x^{2}\left(\cosh\left(x\right)-1\right)}{2+2x\sinh\left(x\right)-2\cosh\left(x\right)-x^{2}}$$

This looks worse than the original limit.

vbxr
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  • What is your definition of $\mathrm{e}^x$? – Xander Henderson Aug 21 '24 at 03:05
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    Guys, please stop answering the question in the comments. Particularly if you are going to provide answers which seem to go against the arbitrary rules established by the question. – Xander Henderson Aug 21 '24 at 03:13
  • To clarify my previous comment, Void Break: you have put a lot of rules on the computation of this limit. My assumption is that you are trying to perform the computation using only very elementary tools (so series solutions, integral solutions, etc are off the table). If this is true, please edit your question to make that more clear. Give us more context about this limit---what are you studying? are methods from complex analysis permissible? integral solutions? Why are you imposing these arbitrary restrictions? – Xander Henderson Aug 21 '24 at 03:16
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    Also, because your limit involves the exponential function, it is necessary (particularly in light of the restrictions you've imposed) to give a definition of this function. The definition I prefer, for example, is in terms of a series, so the most elementary argument I can think of involves series. You clearly have something else in mind, so please edit to question to clarify. – Xander Henderson Aug 21 '24 at 03:19
  • Don't know if it helps much, but using $x\mapsto -x$ symmetry gives a cleaner formula, if you apply it to $1/L$. You may want the denominator to be invariant under the symmetry for easier manipulation? – Jyrki Lahtonen Aug 21 '24 at 03:20
  • Anyway, I'll second Xander's reservations. – Jyrki Lahtonen Aug 21 '24 at 03:23
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    You may need to tighten up the estimates provided here. Again, something that will be easier to do with derivatives. – Jyrki Lahtonen Aug 21 '24 at 03:27
  • Apologies everyone. This is merely a curiosity I have from my introductory calculus course. I don't think it has any true mathematical purpose, I'm just interested in puzzles like these. I have edited my post to add some context. – vbxr Aug 21 '24 at 03:48
  • @AnuragA edited, thank you – vbxr Aug 21 '24 at 06:06
  • @AnneBauval I've edited my post to use the limit definition of e^x now. Sorry about mentioning AnuragA. – vbxr Aug 22 '24 at 07:14
  • In my opinion, you have made a workflow mistake. Based on the posted question, it is quite plausible that (absent L'Hopital's rule) that the problem composer (AKA your teacher) does not know of any alternative approach. So, if I was the problem solver, before attempting to tackle the problem, with graphing also excluded, I would go to the teacher. That is, it is entirely plausible that, from the teacher's perspective, there is no alternative approach that uses elementary calculus. – user2661923 Sep 01 '24 at 05:01
  • Even if it were conclusive to determine the value of the limit, your attempt would not prove that the limit exists. – Anne Bauval Sep 10 '24 at 16:08

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Equivalently, let us prove that $$\lim_{x\to0}\frac{e^x-1-x-x^2/2}{x^2}=0.$$ $$\left(1+\frac xn\right)^n=1+x+\frac{n-1}{2n}x^2+\sum_{k=3}^n\binom nk \left(\frac xn\right)^k$$ hence if $|x|\le1$, $$\left|\left(1+\frac xn\right)^n-1-x-\frac{x^2}2\right|\le\frac{x^2}{2n}+|x|^3\left(\left(1+\frac1n\right)^n-1-1-\frac{n-1}{2n}\right).$$ Letting $n$ tend to $\infty$, we get $$\left|e^x-1-x-\frac{x^2}2\right|\le |x|^3\left(e^1-2.5\right)$$ and the result follows.

Anne Bauval
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