How to show the sequence $x_n = (1 + \frac{x}{n})^{n}$ is bounded above by $e^x$? How to show the sequence $x_n = (1 + \frac{x}{n})^{n}$ is bounded above by $e^x$?
Note: I'm not supposed to be able to use any differentiation techniques if possible. Since we techincally "don't know" it yet.
As can be deduced I am trying to show that the sequence $x_n = (1 + \frac{x}{n})^{n}$ is convergent. I have to arrive at this conclusion using the monotonic convergence theorem. So we are given by definition that $$e^{x} = \lim_{n \rightarrow \infty} \Bigg(1 + \frac{x}{n} \Bigg)^{n}$$
I think I figured out how to show the sequence is montonically increasing. My problem is showing that it is bounded. So one idea I thought of trying to apply was using the binomial theorem:  
$$\Bigg(1 + \frac{x}{n}\Bigg)^{n} = \sum_{k = 0}^{n} \binom{n}{k} \bigg(\frac{x}{n}\Bigg)^{k}$$  and then since this would be some sort of finite quantity, I would compare it to $e^x$:
$$\Bigg(1 + \frac{x}{n}\Bigg)^{n} = \sum_{k = 0}^{n} \binom{n}{k} \bigg(\frac{x}{n}\Bigg)^{k} < e^{x} = \lim_{n \rightarrow \infty} \Bigg(1 + \frac{x}{n} \Bigg)^{n} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots $$
But I can't seem to get the finite binomial expansion in a comparable form.
Questions: 
1) Is this a correct approach ?
2) If it is how can I rewrite the binomial expansion to work in my favor ?
 A: The result is not true for $x<0$. A better approach for $x >0$:  If $x >0$ the $(1+\frac x n)^{n}=e^{n\log (1+x/n)}\leq e^{n\frac x n}=e^{x}$ where I have used the inequality $\log(1+y) \leq y$ for all $y >0$.
A: Hint: If $t>0$ then $$\log(1+t)=\int_1^{1+t}\frac{ds}s<\int_1^{1+t}ds=t.$$
A: For $x\ge 0 $, $(1+\frac{x}{n})^n=1+x+\frac{n(n-1)x^2}{2n^2}+\frac{n(n-1)(n-2)x^3}{6n^3}+...\lt 1+x+\frac{x^2}{2}+\frac{x^3}{6}+...=e^x$  Therefore the term is bounded and increasing with $n$
Essentially you have $\frac{\binom{n}{k}}{n^k}\le \frac{1}{k!}$
A: With calculus and elementary method, that is by MVT for $y=\ln(1+t)$
$$\ln(1+a)-ln(a)=\dfrac{1}{\xi}$$
where $a<\xi<a+1$ which shows
$$\ln(1+\dfrac1a)=\dfrac{1}{\xi}<\dfrac{1}{a}$$
or $\left(1+\dfrac1a\right)^a<e$ now let $a=\dfrac{n}{x}$.
A: If $x>0$, how $$\Bigg(1 + \frac{x}{n}\Bigg)^{n} = \sum_{i = 0}^{n} \binom{n}{i} \bigg(\frac{x}{n}\bigg)^{i}$$ then if $i=k$, the $k-th$ term in this serie is:
$$ \binom{n}{k} \bigg(\frac{x}{n}\bigg)^{k}=\frac{x^k}{k!}\cdot\frac{n!}{(n-k)!n^k}$$ Note that $$\frac{n!}{(n-k)!n^k}\leq1$$ 
Therefore
$$ \binom{n}{k} \bigg(\frac{x}{n}\bigg)^{k}=\frac{x^k}{k!}\cdot\frac{n!}{(n-k)!n^k}\leq \frac{x^k}{k!}$$
A: From your question it appears that the goal here is to justify that the definition $$e^x=\lim_{x\to\infty} \left(1+\frac{x}{n}\right)^n\tag{1}$$ makes sense. Otherwise if one assumes the existence of the above limit the boundedness of the corresponding sequence is automatically guaranteed (the comment by Jack Crawford and my response to it are based on this assumption).
I hope you are familiar with the proof (given in most textbooks) that the sequence $(1+(1/n))^n$ is increasing and bounded above (say by $3$) and hence it converges to a limit conventionally denoted by $e$. If $k$ is a positive integer then we have $$\left(1+\frac{k}{n}\right)^n=\prod_{i=1}^{k}\left(1+\frac{1}{n+i-1}\right)^{n+i-1}\cdot\left(1+\frac{1}{n+i-1}\right)^{1-i}\tag{2}$$ Each term in the product on right side tends to $e\cdot 1=e$ and hence the overall product tends to $e^k$.
What we have proved here is that the sequence in question is convergent if $x$ is a positive integer. To deal with the case when $x$ is positive but not necessarily an integer we need to take a positive integer, say $k$, such that $k>x$ (we may take $k=\lfloor x\rfloor +1$). Then we have $$\left(1+\frac{x}{n}\right)^n\leq \left(1+\frac {k} {n} \right) ^n\tag{3}$$ The right side above converges to $e^k$ and hence is abounded above. Consequently the left side is also bounded above.
