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?
How to evaluate $\lim\limits_{x \rightarrow +\infty}{e^x (e - (1+\frac{1}{x} )^x)}$ without L'Hospital?
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1It is always possible to find a limit without l'Hopital as at some point l'Hopital was proving (hopefully without invoking l'Hopital). – 2012-08-15
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5what we allowed to use? – 2012-08-15
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0You are right. I never watched that. Thanks. – 2012-08-16
2 Answers
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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4So that's why they call it "the *natural* logarithm"! – 2012-08-15
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0Taylor series is much more abvanced approch than L'Hopital rule. So I think this is not a solution or a hint. – 2012-08-15
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0Yes, i try only use "simple limits". – 2012-08-15
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1@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). – 2012-08-15
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3@Norbert: The Taylor series is only "more advanced" in the sense that it's taught later in the curriculum. – 2012-08-15
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0@jonjones: We basically want to know how close $(1+1/x)^x$ and $e$ are, for large $x$. We can get good estimates by using the Binomial Theorem, which is in a sense more elementary, but requires much more attention to detail. "Elementary" arguments can require more mathematical experience than less elementary ones. – 2012-08-15
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0how are taylor series more advanced than l'hopital? if youre working with analytic functions you might as well use series expansions, some people would say it's much more natural – 2012-08-15
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1The 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? – 2012-08-15
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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0$\lim\limits_{x\to\infty} \left(1+\frac1x\right)^x = e$. So the argument doesn't work. – 2013-11-28
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0@DanielFischer, which argument?, in this case $M=e$ – 2013-11-28
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1So if you look at $\lim\limits_{x\to\infty} e^x(e-M)$, you look at $\lim\limits_{x\to\infty} e^x\cdot 0$. – 2013-11-28
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0@DanielFischer, ahh I see. thanks for pointing out that, I will try to find better answer. – 2013-11-28