How to prove that $\lim_{n\to\infty}\big(1+1/n\big)^n$ is equal to e How do I prove the following limit without using the derivative. 
$$\lim_{n\to\infty}\big(1+1/n\big)^n$$
I have tried using the Binomial theorem but I haven't got too far. I have proved that the limit is between 2 and 3 and that it is convergent. Also if I logaritmate the whole equation I can't get anywhere without applying the derivative. And it doesn't get me anywhere if I try the Epsilon-delta proof.
$$\ln(y) = \ln\left(\lim_{n\to\infty}\big(1+1/n\big)^n\right)$$
 A: What you have written is often taken as the definition of $e$.  You know that it converges, and the number it converges to is what we call $e$.
Another formula, useful in approximating $e$ (because it converges quickly), is this one:
$$e = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots = 1 + 1 + \frac{1}{2} + \frac{1}{6}+\cdots$$
To show that this is equal to what you've written, we should start with the binomial theorem.  The $k$-th term of $(1+1/n)^n$ is ${n\choose k} n^{-k}$, or $\frac{n(n-1)\cdots(n-k+1)}{k! n^k} = \frac{1\cdot(1-1/n)\cdots(1-(k+1)/n)}{k!}$.  If $k$ remains fixed, and $n\to \infty$, this is just $\frac{1}{k!}$.
This is not quite a proof, but it's a good start for comparing the two expressions.  Once you know the above power series, you can compute that $e=2.7182818284\ldots$ or whatever you like.
A: Let $E(n)=(1+1/n)^n.$ Notice that $$\ln(E(n))=\frac{\ln(1+1/n)}{1/n}.$$
Therefore the limit as $n\to \infty$ of $\ln E(n))$ is
$$
\lim_{1/n\to 0} \frac{\ln(1+1/n)-\ln(1)}{1/n}
$$
which is the derivative of $\ln(x)$ near $1$ and therefore equals $1$.  Using continuity of the natural logarithm, this shows that 
$$
\ln\lim_{n\to\infty}E(n)=1.
$$
Therefore the limit must be $e$ where $e$ is the unique number whose natural log is $1$. 
A: $$
\begin{align}
&\lim_{n\to\infty}(1+\frac 1n)^n\\
=&\lim_{n\to\infty} e^{n\log(1+1/n)}\\
=&e^{\lim_{n\to\infty}n\log(1+1/n)}\\
=&e^{\lim_{t\to 0}\log(1+t)/t}\\
=&e^{1/1}\\
=&e
\end{align}
$$
Explanations:
$1$.Given
$2$.$e^{\log x}=x$
$3$.$\lim a^x=a^{\lim x}$
$4$.$ \,t=\frac 1n$
$5$.Limit Evaluation
$6$.Simplification
A: Before at all you have to prove that :
$s_n=(1+\frac{1}{n})^n$ converges into a finite value. 
How? 
Firstly prove that $s_n$ is monotone-increasing sequence. 
Secondly prove that is a upper bounded sequence.
If this 2 statements are true then $s_n$ converges into a finite value (important theorem)
and then we call it e
After We wonder how to aproximate e. But that question was answered by the others above.
