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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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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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@DanielFischer, ahh I see. thanks for pointing out that, I will try to find better answer. – 2013-11-28