# How to evaluate $\lim\limits_{x \rightarrow +\infty}{e^x (e - (1+\frac{1}{x} )^x)}$ without L'Hospital?

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?

-
It is always possible to find a limit without l'Hopital as at some point l'Hopital was proving (hopefully without invoking l'Hopital). – Fabian Aug 15 '12 at 20:49
what we allowed to use? – Norbert Aug 15 '12 at 20:55
You are right. I never watched that. Thanks. – jon jones Aug 16 '12 at 0:17

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.

-
So that's why they call it "the natural logarithm"! – Asaf Karagila Aug 15 '12 at 21:11
Taylor series is much more abvanced approch than L'Hopital rule. So I think this is not a solution or a hint. – Norbert Aug 15 '12 at 21:11
@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). – André Nicolas Aug 15 '12 at 21:14
@Norbert: The Taylor series is only "more advanced" in the sense that it's taught later in the curriculum. – Hurkyl Aug 15 '12 at 21:18
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? – GEdgar Aug 15 '12 at 23:35

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.

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

-
$\lim\limits_{x\to\infty} \left(1+\frac1x\right)^x = e$. So the argument doesn't work. – Daniel Fischer Nov 28 '13 at 18:57
@DanielFischer, which argument?, in this case $M=e$ – chatur Nov 28 '13 at 19:00
So if you look at $\lim\limits_{x\to\infty} e^x(e-M)$, you look at $\lim\limits_{x\to\infty} e^x\cdot 0$. – Daniel Fischer Nov 28 '13 at 19:02
@DanielFischer, ahh I see. thanks for pointing out that, I will try to find better answer. – chatur Nov 28 '13 at 19:14