How to find $\lim\limits_{n\to \infty }(1 + \frac{1}{n})^n$ $\lim _{n\to \infty }\left(1 + \frac{1}{n}\right)^n$
How do I get started with this one? Variable substitution would be one way, but our lecturer hasn't covered that yet, so there should be some other way.
Usually with these kinds of limits we modify the function so that it resembles one of the standard limits that we may use without proving them.
 A: We only need the fact that $f(x)=e^x - 1- x >0, x\neq 0$
Firstly $e^{1/n} > 1 + \frac{1}{n}$, i.e.
$$(1+\frac{1}{n})^n < e$$
Secondly $e^{-\dfrac{1}{n+1}} > 1 - \frac{1}{n+1}$, i.e.
$$\dfrac{1}{e^{\dfrac{1}{n+1}}} > 1 - \dfrac{1}{n+1}$$
thus $$e^{\dfrac{1}{n+1}} < \dfrac{1}{1 - \dfrac{1}{n+1}} = \dfrac{n+1}{n} = 1 + \dfrac{1}{n}$$
so $$ (1 + \frac{1}{n})^n > e^{\dfrac{n}{n+1}}$$
In summary $$ e^{\dfrac{n}{n+1}}< (1+\frac{1}{n})^n < e$$
Sending $n$ to infinity gives that the limit is equal to $e$
A: I know of a way to do this with the use of binomial theorem. You have 
$$\left(1 + \frac{1}{n}\right)^n=\binom{n}{0}+\binom{n}{1}\frac{1}{n}+\binom{n}{2}\frac{1}{n^2}+\binom{n}{3}\frac{1}{n^3}+....$$ 
$$=1+1+\frac{n(n-1)}{2!\cdot n^2}+\frac{n(n-1)(n-2)}{3!\cdot n^3}+...$$ 
$$=1+1+\frac{1-\frac{1}{n}}{2!}+\frac{1-\frac{3}{n}+\frac{2}{n^2}}{3!}+...$$
Now as $n$ goes towards infinity, some terms in the numerators vanish and what remains is this:
$$\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...$$
Which we know to be $e$.
A: It depends how you define $e$. Many texts do indeed define $\lim_{n \to \infty} (1 + \frac{1}{n})^{n}$ as $e,$ but to do that, it is usual to prove that the sequence is increasing, and is bounded above. Any increasing sequence of real numbers which is bounded above tends to its least upper bound. Once it is established that the limit exists, it is reasonable to give it a name, and some texts choose to define $e$ that way.
If you already know the Taylor series definition of $e$ or $e^{x}$ then it is easy to check that 
$(1 + \frac{1}{n})^{n} = 1 + \sum_{k=1}^{n} \frac{1}{k!}\prod_{j=1}^{k-1}(1- \frac{j}{n})$
which is always less than $\sum_{k=0}^{\infty} \frac{1}{k!}.$ Proving the inequality the other way round (in the limit) is more difficult.
If the $\log$ function is used to define $e$ via $\log(x) = \int_{1}^{x} \frac{1}{t} dt$
and $\log (e) = 1,$ then  we have $ n \log(1 + \frac{1}{n}) < n.\frac{1}{n} =1,$ so that $(1+\frac{1}{n})^{n} < e$ for all $n.$ On the other hand, $\log(1 + \frac{1}{n}) > \frac{1}{n} - \frac{1}{2n^{2}},$ so $n \log(1 + \frac{1}{n}) > 1 - \frac{1}{2n}$ and thus $n \log(1 + \frac{1}{n}) \to 1$ as $n \to \infty$.
A: it is known that the limit exists and is $e$. here is one way show that the limit exists and is between $2$ and $3.$ we will use the facts $(1 + \frac{1}{n})^n$ is an increasing sequence and it is bounded above by $3.$  with that the cliam will be established.
that $a_n=(1 + \frac{1}{n})^n$ is increasing follows from
$\dfrac{a_{n+1}}{a_n} = \left(\dfrac{n+2}{n+1}\right)\left( \dfrac{n^2 + 2n}{n^2 + n}\right)^n > 1$ for all $n \ge 1.$
 and the boudedness follows from
$\begin{align}(1 + \frac{1}{n})^n &= 1 + 1 + \dfrac{(1-1/n)}{2} + 
\dfrac{(1-1/n)(1-2/n)}{2*3} + \dfrac{(1-1/n)(1-2/n)(1-3/n)}{2*3*4}+ \cdots\\
& \le 1 + 1 + \dfrac{1}{2} + \dfrac{1}{2*3} + \dfrac{1}{2*3*4} + \cdots\\
& < 1 + 1 + \dfrac{1}{2} + \dfrac{1}{2*2} + \dfrac{1}{2*2*2} + \cdots = 3\\
\end{align}$
