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I have the limit $$ \lim_{x\to0_+} \frac{e^x - x e^x - 1}{\left(e^x - 1 \right)^2} $$ which I need to compute without L'Hôpital's rule.

(The result is $-\frac{1}{2}$ with L'Hôpital's rule).

Thanks.

AccidentalFourierTransform
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Hamza Lotfi
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  • Comment/hint since I don't want to take the time to write an answer: write out the power series expansions for the numerator and denominator, start the formal long division and look at the constant term. – Ethan Bolker Aug 11 '17 at 13:52
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    Why are there so many questions here about limits without using l'Hopital's rule? Is that something students typically do? It's very easy to prove; and if teachers didn't want students to use it for some reason, it would make more sense to design problems where it wasn't immediately applicable. – anomaly Aug 12 '17 at 00:16
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    @anomaly: if you see various limit questions and their answers on this site you will be convinced that most students simply don't have an exact idea of the hypotheses and conclusions of the theorem called "L'Hospital's Rule". And I doubt if these students really know the proof. In fact most students believe that L'Hospital's Rule is a thumb rule which says that if plugging does not work then try differentiation and plugging and perhaps this might be needed multiple times for some problems. – Paramanand Singh Aug 12 '17 at 10:30
  • @anomaly : you may look at this answer (https://math.stackexchange.com/a/2382324/72031) which has a subtle flaw in application of L'Hospital's Rule which is difficult to detect (but easy to fix). Also see the comments where I describe the flaw. – Paramanand Singh Aug 12 '17 at 10:36
  • This seems like a homework question to me. – Clonkex Aug 12 '17 at 12:01
  • @ParamanandSingh: I've actually seen a few (but only a few) such questions here where l'Hopital's rule isn't directly applicable. Those would be much better exercises to give students rather than artifically removing a simple, easily proved tool from their toolbox. – anomaly Aug 12 '17 at 16:17

4 Answers4

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We write $$\frac {e^x-xe^x-1}{x^2}\color {red}{\frac {x^2}{(e^x-1)^2}}. $$

Let us begin by

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

put $x^2=t $ and $$f (t )=e^{\sqrt {t}}-\sqrt {t}e^{\sqrt {t}}$$

then $$L=\lim_{0^+}\frac {f (t)-f (0)}{t} =f'(0) $$ and since $$f'(t)=-\frac {1}{2}e^{\sqrt {t}} $$ we have

$$f'(0)=-\frac {1}{2} $$

the $\color {red}{red }$ fraction goes to $1$.

the final result is $-\frac {1}{2} $.

einpoklum
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hamam_Abdallah
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    Would have been better to just apply Taylor series in your second expression (whose limit is same as that of original expression as you have shown. Trying to recognize that limit as a derivative of a complicated function is not so obvious here. Moreover once you differentiate, then you have already crossed the algebra of limits and it does not matter whether you use differentiation for L'Hospital's Rule or to calculate Taylor series coefficients and out of all this Taylor series is the simplest. – Paramanand Singh Aug 11 '17 at 18:14
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    Taylor series is a generalization of MVT which come from derivative. – hamam_Abdallah Aug 11 '17 at 20:37
10

It is also simple to look at Taylor expansions around $0$ up to at least second order $$e^x = 1 + x + \frac{1}{2}x^2... $$ then $$\frac{e^x - xe^x - 1}{(e^x-1)^2} \approx \frac{1 + x + \frac{1}{2}x^2 - x - x^2 - \frac{1}{2}x^3 - 1} {(x+ \frac{1}{2}x^2)^2} = \frac{ - \frac{1}{2}x^2 - \frac{1}{2}x^3}{x^2 + \frac{1}{4}x^4 + x^3} $$ and now by looking at the lowest order terms, the conclusion that $$\lim_{x \to 0^+} = -\frac{1}{2}$$ follows quick

An aedonist
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$\lim \limits_{x \to0+}$ $\frac{e^x-xe^x-1}{\left(e^x-1\right)^2}\cdot\frac{x^2}{x^2}$= $\lim \limits_{x \to0+}$ $\frac{e^x-x+x-xe^x-1}{x^2}\cdot\lim \limits_{x \to0+}$ $\frac{x^2}{\left(e^x-1\right)^2}$

= $\lim\limits_{x\to 0^{+}}\frac{e^x-x+x-xe^x-1}{x^2}\cdot\lim \limits_{x \to0+}\left[\frac{x}{\left(e^x-1\right)}\right]^2$ =$\lim\limits_{x\to 0^{+}}\frac{e^x-x-1+x-xe^x}{x^2}\cdot1$

=$\lim \limits_{x \to0+}\frac{e^x-x-1}{x^2}-\lim \limits_{x \to0+}\frac{x\left(e^x-1\right)}{x^2}$ =$\frac{1}{2}-\lim \limits_{x \to0+}\frac{e^x-1}{x}$ =$\frac{1}{2}-1=-\frac{1}{2}$

note $\lim \limits_{x \to0+}\frac{e^x-x-1}{x^2}=\frac{1}{2}$ can be proved without Lohspital rule or series

Paramanand Singh
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Hussien Mohamed
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  • Apart from the minor $\mathrm\LaTeX $ issues (which I am fixing with minor edit) this is a very good answer and exactly the sort of answer demanded (without L'Hospital's Rule). Wonder why it was severely downvoted! +1 from my end. – Paramanand Singh Aug 11 '17 at 18:18
  • @ParamanandSingh: Maybe because the font is a bit small? Anyway +1. It's not as "fancy" as the top-voted answer, but like you said, this is what teachers would expect basically (well, either this or Taylor expansion). – einpoklum Aug 12 '17 at 08:48
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Let's give it one more shot even though there are some good answers. We can proceed as follows \begin{align} L&=\lim_{x\to 0^{+}}\frac{e^{x}-xe^{x}-1}{(e^{x}-1)^{2}}\notag\\ &=\lim_{x\to 0^{+}}\frac{e^{x}-xe^{x}-1}{x^{2}}\cdot\left(\frac{x}{e^{x}-1}\right)^{2}\notag\\ &= \lim_{t\to 0^{-}}\frac{e^{-t}+te^{-t}-1}{t^{2}}\notag\\ &=-\lim_{t\to 0^{-}}\frac{e^{t}-t-1}{t^{2}}\cdot\frac{1}{e^{t}}\notag\\ &=-\frac{1}{2}\text{ (via Taylor series)} \notag \end{align}

Paramanand Singh
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