Sets and supremum Please I need help solving this question:
$E(x)= \left \{ \left(1+ \frac{x}{n} \right)^n : n \in \mathbb{N} \right \}$.
Let $a(x) = \sup E(x)$ (least upper bound).
Prove that $a(x) < a(y)$ if $0 < x < y$.
I already proved that E(x) has no largest element and is bounded.
But I'm having trouble with the above proof. Any help please?
 A: As others have shown, there is a lot one can do if one already knows something about real analysis. However, some things are relatively easy even without that knowledge.
As has been remarked, for $0 < x < y$ we have for each $n \in \mathbb{N}$
$$1+\frac{x}n < 1+\frac{y}n$$
and hence
$$\left(1+\frac{x}n\right)^n < \left(1+\frac{y}n\right)^n.$$
Since $a(y)$ is an upper bound of $E(y)$, we get for each $n \in \mathbb{N}$
$$\left(1+\frac{x}n\right)^n < \left(1+\frac{y}n\right)^n \le a(y)$$
that means $a(y)$ is also an upper bound of $E(x)$, so we get
$$a(x) \le a(y)$$
since $a(x)$ is the least upper bound of $E(x)$.
The real problem IMO is showing that the inequality is actually strict, that means $a(x) \neq a(y)$.
To see that, start with a bit of simply algebra:
$$\left(1+\frac{y}n\right) = \left(1+\frac{x}n\right)\left(1+\frac{y-x}{x+n}\right)$$.
From this of course it follows that 
$$\left(1+\frac{y}n\right)^n = \left(1+\frac{x}n\right)^n\left(1+\frac{y-x}{x+n}\right)^n$$.
We want to show that there exists an $f > 1$ such that 
$$\left(1+\frac{y-x}{x+n}\right)^n \ge f\text{ for all }n \in \mathbb{N}. \tag{1}\label{eq1}$$ 
That $f$ of course depends on $x,y$, but since we don't vary those variables, that doesn't matter. The important part is that $f$ does not depend on $n$.
Once we know this, we have 
$$\left(1+\frac{y}n\right)^n \ge f\left(1+\frac{x}n\right)^n$$
and since $\left(1+\frac{x}n\right)^n$ gets as near as desired to $a(x)$ ($a(x)$ is least upper bound of $E(x)$), it follows that $\left(1+\frac{y}n\right)^n$ has a least upper bound that is at least $fa(x)$, and since $f>1$ we cannot have $a(x)=a(y)$.
Let's prove $\eqref{eq1}$:
First let's restate the obvious: For all $n$ we have $\left(1+\frac{y-x}{x+n}\right)^n > 1$.
There are only finintely many values of $n$ with $n < x$, so if we set $f_1$ equal to the minimum of $\left(1+\frac{y-x}{x+n}\right)^n$ for $n < x$, we have, for $n < x$:
$\left(1+\frac{y-x}{x+n}\right)^n \ge f_1 > 1$
For $n \ge x$ we have $n+x \le 2n$, hence $\frac{y-x}{n+x} \ge \frac{y-x}{2n}$ (all terms are positive). From this follows
$\left(1+\frac{y-x}{n+x}\right)^n \ge \left(1+\frac{y-x}{2n}\right)^n \ge 1+\frac{y-x}2$
where the last inequality is simply using the binomial expansion of the term on the left and stopping after the linear term (the remaining terms are all non-negative).
Again we have $f_2=1+\frac{y-x}2 > 1$ so our $f$ becomes
$f=\min (f_1,f_2) > 1$
which concludes the proof.
A: Hint: $a(x)=e^x$.  And $e^x$ is strictly increasing. 
A: You also can prove that the sequence $E(x)$ is strictly increasing in $n$ for large $n$ (take the derivative with respect to $n$), which in turn tells you that $a(x)$ is the limit of the sequence, not just its supremum.
Do you know L'Hopital's Rule?  Since we're dealing with limits, we can use it to determine $\lim_{n \to \infty} \ln(E(x))$.  Since $\ln$ is continuous, $\ln a(x) = \ln (\lim_{n \to  \infty} E(x)) = \lim_{n \to \infty} \ln(E(x))$, so $a(x) = e^{\lim_{n \to \infty} \ln(E(x))}$.
