Show that $2\leq \left(1+\frac{1}{x}\right)^{x}<2.8$ I was asked by a junior of mine to explain to her how the following is true:
For all $x\geq 1$,
\begin{eqnarray}
2\leq \left(1+\frac{1}{x}\right)^{x}<2.8.
\end{eqnarray}
I know that $\left(1+\frac{1}{x}\right)^{x}\to \text{e}\approx2.7$ as $x\to \infty$. However, she does not have the requisite background to understand limits.
Can someone point me out to a more preliminary level at which this problem can be dealt?
 A: 
I thought it might be useful to present a way forward in which $x$ is any positive integer.  To that end we proceed.


In THIS ANSWER, I showed using only Bernoulli's Inequality that $\left(1+\frac1n \right)^n$ is monotonically increasing for integer values of $n\ge 1$.
From the binomial theorem, we have for $n\ge 1$
$$\begin{align}
\left(1+\frac1n \right)^n&=1+1+\frac1{2!} \left(1-\frac1n\right)+\frac1{3!}\left(1-\frac1n\right)\left(1-\frac2{n}\right)+\cdots \frac1{n!}\left(1-\frac{n-1}{n}\right)\\\\
&\le \sum_{k=0}^n\frac1{k!}\\\\
&\le \sum_{k=0}^\infty\frac1{k!}\\\\
&= 1+1+\frac12+\frac16+\sum_{k=4}^\infty \frac{1}{k!}\\\\
&\le 2\frac23+\sum_{k=4}^\infty \frac{1}{2^k}\\\\
&=2\frac23+\frac18\\\\
&=2\frac{19}{24}\\\\
&<2.8
\end{align}$$
From the monotonicity, the lower bound is evidently $2$ (i.e., the term of interest is greater than its value at $n=1$).  Putting it together reveals 
$$2\le \left(1+\frac1n\right)^n<2.8$$
for integer values of $n\ge 1$.
A: Let $f(x)=(1+\frac1x)^x$ for $x\geq 1$.
$g(x)=\ln(f(x))=x\ln(\frac{1+x}{x})$
$g'(x)=\ln(1+x)-\ln(x)-\frac{1}{1+x}$
$=\frac{1}{c_x}-\frac{1}{1+x}\;\;$ by MVT.
$>0\;\;$ since $\;\;x<c_x<1+x$.
thus
$g\;$ and $\;f$ are increasing and
$\forall x\geq 1$
$$f(1)=2<f(x)<\lim_{x\to+\infty}f(x)=e<2.8$$
A: This is partial answer since you need to be able to show that given function is increasing/decreasing, for which you might need to use derivations (but for integer numbers not necessarily so).
So assuming you are able to show that $\left(1+\frac{1}{x}\right)^x$ is increasing for $x \geq 1$, lower bounds follows, since you have
$$2 = \left(1+\frac{1}{1}\right)^1 \leq \left(1+\frac{1}{x}\right)^x.$$
On the other hand if you are able to show that $\left(1+\frac{1}{x}\right)^{x+\frac{1}{2}}$ is decreasing for $x\geq 1$, the upper bound follows for $x \geq 2$ since
$$\left(1+\frac{1}{x}\right)^x \leq \left(1+\frac{1}{x}\right)^{x+\frac{1}{2}} \leq \left(1+\frac{1}{2}\right)^{2+\frac{1}{2}} \approx 2.775$$
A: Let $h(x)=\ln(1+x)-x\;\;$ for $x\geq0$.
$$h'(x)=\frac{1}{1+x}-1$$
$$=\frac{ -x}{  1+x   }\leq 0$$
thus 
$$(\forall x>0)\; \;\; h(x)\leq h(0)$$
$\implies$
$$(\forall x>0)\;\;\ln(1+x)\leq x$$
or
$$(\forall x>0)\;\; \ln(1+\frac 1x)\leq \frac 1x$$
$\implies$
$$(\forall x>0)\;\; x\ln(1+\frac 1x)\leq \ln(e)$$
And finally
$$(\forall x\geq 1)\;\; 2\leq (1+\frac 1x)^x\leq e$$
