5

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?

Curious student
  • 6,583
jon jones
  • 1,178

2 Answers2

6

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.

André Nicolas
  • 507,029
  • 4
    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
  • Yes, i try only use "simple limits". – jon jones Aug 15 '12 at 21:14
  • 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). – André Nicolas Aug 15 '12 at 21:14
  • 3
    @Norbert: The Taylor series is only "more advanced" in the sense that it's taught later in the curriculum. –  Aug 15 '12 at 21:18
  • @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. – André Nicolas Aug 15 '12 at 21:19
  • how 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 – butt Aug 15 '12 at 22:20
  • 1
    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
-1

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.

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

chatur
  • 1,132