How is $\displaystyle\lim_{n\to\infty}{(1+1/n)^n} = e$? How is $\displaystyle\lim_{n\to\infty}{(1+1/n)^n} = e$? Please help by letting me know why in a very simple way. Thank you.
 A: First you show that $a_n=(1+1/n)^n$ is increasing, $b_n=(1+1/n)^{n+1}$ is decreasing. Since $a_n\leq b_n,\ \forall n,$ both $a_n$ and $b_n$ converge, oviously to the same limit. Then you define $e:=\lim_{n\to\infty} a_n.$
A: This is the definition of $e$. I can say $\text{zafer}=\pi/2$, and this is a definition. There is no proof of this, because this is how I define $\text{zafer}$. 
A: I guess, what you really want to know is: How can I see, that the limit
$$e = \lim_{n\rightarrow \infty}(1+\frac{1}{n})^n$$
has the known properties like $\frac{d}{dx}e^x = e^x$.
First of all, the function $e^x$ can be easily expanded as an infinity series $e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$. Knowing, that this series is absolute convergent, on can simply derive each summand and see that $\frac{d}{dx}e^x = e^x$.
If you set $x=1$ you get a series, which is mentioned here as alternative definition of $e$. The prove that this definition is equivalent to yours can be found here.
Hope that answers your question.
A: Here is the basic fact that I think you are reaching for:
Let $T_n=(1+1/n)^n$. Let $S_n=1+1+1/2!+\ldots+1/n!$. Then the limits as $n$ tend to infinity of $T_n$ and $S_n$ (call them $T$ and $S$ repectively) exist and are equal.
Existence of $S$: $S_n$ is increasing and bounded above by 
\begin{equation*}
1+1+\frac{1}{2}+\frac{1}{2\cdot 2}+\frac{1}{2\cdot 2\cdot 2}+\ldots=3
\end{equation*}
Existence of $T$: Expand $T_n$ by the Binomial Theorem and rearrange to obtain
\begin{align*}
T_n=1+1&+(1-1/n)\frac{1}{2!}+(1-1/n)(1-2/n)\frac{1}{3!}+\ldots\\
&+(1-1/n)(1-2/n)\cdots(1-(n-1)/n)\frac{1}{n!}.
\end{align*}
A term by term comparison between $T_n$ and $T_{n+1}$ shows that $T_n$ is increasing. A term-by-term comparison between $T_n$ and $S_n$ shows that $T_{n}\le S_n$. Since $S_n\le 3$ it follows that $T_n$ converges.
Proof that $T=S$: Since $S_n\le T_n$ it suffices to show that $T\ge S$. 
Fix $m<n$ and split the expression for $T_n$ given above into two parts
 $$T_n=P_{m,n}+Q_{m,n},$$ where
\begin{align*}
P_{m,n}=&1+1+(1-1/n)\frac{1}{2!}+(1-1/n)(1-2/n)\frac{1}{3!}+\ldots\\
&+(1-1/n)(1-2/n)\cdots(1-(m-1)/n)\frac{1}{m!}
\end{align*}
and 
\begin{align*}
Q_{m,n}=&(1-1/n)(1-2/n)\cdots(1-m/n)\frac{1}{(m+1)!}+\ldots\\
&+(1-1/n)(1-2/n)\cdots(1-(n-1)/n)\frac{1}{n!}.
\end{align*}
Note that  $Q_{m,n}\ge 0$. Note that as $n$ tends to infinity $P_{m,n}$ tends to $S_m$.
Therefore, taking limits as $n\to\infty$ of both sides of the equation 
$T_n=P_{m,n}+Q_{m,n}$, we obtain
$$T=S_m+\lim_{n\to\infty}Q_{m,n}\ge S_m.$$ Since $m$ is arbitrary, it follows that $T\ge S$.
Crystal's Algebra Vol 2 is a wonderful source for such seat-of-the pants proofs.
A: Consider $$\lim_{n\rightarrow \infty}(1+{1\over n})^n.$$ This limit has the indeterminate form $1^\infty$. Let $y=(1+{1\over n})^n$. Taking the natural logarithm of both sides of the equation and simplifying using the rules of logarithms we obtain $\ln(y)=n\ln(1+{1\over n})$. The $$\lim_{n\rightarrow \infty} \ln(y)=\lim_{n\rightarrow \infty}n\ln(1+{1\over n})$$ which has the indeterminate form $\infty\cdot 0$. We can rewrite the right-hand side limit as $$\lim_{n\rightarrow \infty}\ln(y)={\lim_{n\rightarrow \infty}={\ln(1+{1\over n})\over {1\over n}}}$$ which has the indeterminate form ${0\over 0}$. Using L'Hospital's Rule we see that $$\lim_{n\rightarrow \infty}\ln(y)=\lim_{n\rightarrow \infty}{{1\over (1+{1\over n})}\cdot {-1\over n^2}\over {-1\over n^2}}.$$ This simplifies to $$\lim_{n\rightarrow \infty} \ln(y)=\lim_{n\rightarrow \infty}{1\over 1+{1\over n}}=1.$$ So far we have computed the limit of $\ln(y)$, what we really want is the limit of $y$. We know that $y=e^{\ln(y)}$. So $$\lim_{n\rightarrow \infty}(1+{1\over n})^n=\lim_{n\rightarrow \infty} y=\lim_{n\rightarrow \infty} e^{\ln(y)}=e^1=e.$$ Thus $$\lim_{n\rightarrow \infty} (1+{1\over n})^n=e.$$
A: Firstly, you should note that as $n\rightarrow±\infty$, our expression becomes $(\rightarrow 1)^\infty$. Although an exact 1 raised to any number is 1 itself, our expression isn't exactly 1. It's tending to 1. So we can't compute the value by simply plugging in the value for n. That isn't solving a limit.
Let's approach it in another way:
Let $f(x) = (1 + \frac{1}{x})$
and $g(x) = x$
What you want to find out:
$$\lim_{x\to a} f(x)^{g(x)}\\$$
$$\Large = \lim_{x\to a} e^{\ln f(x)^{g(x)}}$$
($\because e^{\ln x} = x)$
$$\Large = \lim_{x\to a} e^{g(x).\ln f(x)}$$
($\because \ln a^b = b\ln a)$  
The index part is getting very tiny and squeezed up making it difficult to read - which is why I'm only solving the index below:  
\begin{eqnarray}
Let L &=& g(x).\ln f(x)\\
 &=& g(x).\ln(1 + f(x) - 1)\\
 &=& g(x).\frac{\ln\bigg(1 + \big(f(x) - 1\big)\bigg)}{\big(f(x) - 1\big)}.(f(x) - 1)\\
\end{eqnarray}
We know that $f(x) - 1 \to0$ $(\because f(x) \to1)$; and that:
$$\lim_{x\to a} \frac{\ln(1 + x)}{x} = 1$$
So we get:
$$L =\lim_{x\to a} g(x)\cdot(f(x) - 1)$$
Hence, your original expression is:
$$\Large e^\left({\lim\limits_{x\to a} g(x)\cdot(f(x) - 1)}\right)$$
Substituting the values for f(x) and g(x):
$$\Large e^\left({\lim\limits_{x\to \infty} g(x)\cdot(f(x) - 1)}\right)$$
$$=\Large e^\left({\lim\limits_{x\to \infty} x\cdot(1 + \frac{1}{x} - 1)}\right)$$
$$=\Large e^\left({\lim\limits_{x\to \infty} x\cdot(\frac{1}{x})}\right)$$
$$=\Large e^1$$
$$=e$$
A: Let $e_n=(1+\frac{1}{n})^n, n>0$. 
1) $e_n$ is increasing: $$e_n<e_{n+1}<=>(1+\frac{1}{n})^n <(1+\frac{1}{n})^{n+1}<=>\sqrt[n+1]{(1+1/n)^n}<1+\frac{1}{n+1}.$$ 
Last inequality is demonstrated by applying the AM-GM numbers:
$$1+\frac{1}{n},1+\frac{1}{n},...,1+\frac{1}{n}, 1.$$
Otherwise, compare term by term the sums of $e_n$ and  $e_{n+1}$ developing with Newton's binomial.
2)$e_n$ is bounded by 3:
By mathematical induction shows that $$(1+\frac{1}{n})^k < 1+\frac{k}{n}+\frac{k^2}{n^2}$$
for $n>0$ and $k>0$ and then put $k=n.$
Otherwise, increase the sum obtained by developing the binomial terms of Newton to the sum
$$1+1+\frac{1}{1.2}+ \frac{1}{2.3}+...+\frac{1}{(n-1)n}<3.$$
Of  1) and 2) result that $e_n$  converges and the limit is denoted $e$.
