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Using several times L'Hospital Rule I got $$\lim_{x \rightarrow +\infty}{e^x \left (e - \left(1+\dfrac{1}{x}\right )^x\right)} = +\infty.$$ Is it possible find this limit without L'Hospital?

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    $\begingroup$ It is always possible to find a limit without l'Hopital as at some point l'Hopital was proving (hopefully without invoking l'Hopital). $\endgroup$
    – Fabian
    Aug 15, 2012 at 20:49
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    $\begingroup$ what we allowed to use? $\endgroup$
    – Norbert
    Aug 15, 2012 at 20:55
  • $\begingroup$ You are right. I never watched that. Thanks. $\endgroup$
    – jon jones
    Aug 16, 2012 at 0:17

2 Answers 2

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The natural thing to do is to look at the logarithm of $\left(1+\frac{1}{x}\right)^x$, that is, at $x\log\left(1+\frac{1}{x}\right)$. Use the series $$\log(1+t)=t-\frac{t^2}{2}+\frac{t^3}{3}-\frac{t^4}{4}+\cdots.$$ From this we can obtain good estimates of the difference between $e$ and $(1+1/x)^x$ when $x$ is large. For the calculation, the series expansion of $e^t$ is useful.

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    $\begingroup$ So that's why they call it "the natural logarithm"! $\endgroup$
    – Asaf Karagila
    Aug 15, 2012 at 21:11
  • $\begingroup$ Taylor series is much more abvanced approch than L'Hopital rule. So I think this is not a solution or a hint. $\endgroup$
    – Norbert
    Aug 15, 2012 at 21:11
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    $\begingroup$ @AsafKaragila: We may be cursed with a similar sense of humour. I thought of but this time resisted writing, instead of logarithm, natural logarithm (with the emphasis). $\endgroup$ Aug 15, 2012 at 21:14
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    $\begingroup$ @Norbert: The Taylor series is only "more advanced" in the sense that it's taught later in the curriculum. $\endgroup$
    – user14972
    Aug 15, 2012 at 21:18
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    $\begingroup$ The OP only said "without l'Hospital", not "without l'Hospital or Taylor series". AND he did not include label "homework" so why not use any method? $\endgroup$
    – GEdgar
    Aug 15, 2012 at 23:35
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We need to prove that, $$\lim_{x \rightarrow +\infty}{e^x \left (e - \left(1+\dfrac{1}{x}\right )^x\right)} = +\infty.$$

consider $$\lim_{x \rightarrow +\infty}{e^x \left (e - M\right)} $$

where, $$M = \left(1+\dfrac{1}{x}\right )^x$$

if we prove that $M$ has a finite limit, we are done.

Note that,

1. M is increasing function of x
2. M is bounded above

first one you can prove as an exercise, for second $$M = \left(1+\dfrac{1}{x}\right )^x = \left(\left(1+\dfrac{1}{x}\right )^{x/k} \right)^k < \left(\frac{1}{1-\frac{1}{x} \cdot\frac{x}{k}}\right)^k = \left ( \frac{1}{\left(1-\frac{1}{k}\right)^k}\right)$$

so that, $$M< \frac{1}{\left(1-\frac{1}{k}\right)^k}$$ for any whole k.

$$\lim_{x \rightarrow +\infty}{e^x \left (e - M\right)} = \lim_{x \rightarrow +\infty}{e^x L} = +\infty $$

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  • $\begingroup$ $\lim\limits_{x\to\infty} \left(1+\frac1x\right)^x = e$. So the argument doesn't work. $\endgroup$ Nov 28, 2013 at 18:57
  • $\begingroup$ @DanielFischer, which argument?, in this case $M=e$ $\endgroup$
    – chatur
    Nov 28, 2013 at 19:00
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    $\begingroup$ So if you look at $\lim\limits_{x\to\infty} e^x(e-M)$, you look at $\lim\limits_{x\to\infty} e^x\cdot 0$. $\endgroup$ Nov 28, 2013 at 19:02
  • $\begingroup$ @DanielFischer, ahh I see. thanks for pointing out that, I will try to find better answer. $\endgroup$
    – chatur
    Nov 28, 2013 at 19:14

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