prove $\lim\limits_{x \to \infty}(1+ 1/n)^n = e$ without using l'Hospitals rule I did $y=(1+1/n)^n)$ so $ln(y)=n*ln(1+1/n)$ so then $ln(y)=Ln(1+1/n)/(1/n)$ which sith l'hospitals rule becomes $ln(y)=(-1/(n^2 + n))/(-1/n^2)$ which simplifies to $n/(n+1)$ which using l'hospitals rule is $1$ so the answer is e. how do you solve this without l'hospitals rule
Let $f(x) = \ln(x)$ and then go through the process of evaluating $f'(1)$ using the formal definition of the derivative, noting ahead of time that the actual value of $f'(1)$ is $1$: 

$$ 1 = f'(1) = \lim_{h \to 0}\frac{f(1+h)-f(1)}{h} = \lim_{x \to 0}\frac{f(1+x)-f(1)}{x} = \lim_{x \to 0}\ln(1 + x)^{1/x}; $$

and so, 

$$ e^{1} = e^{\lim_{x \to 0}\ln(1+x)^{1/x}} = \lim_{x \to 0}(1+x)^{1/x}. \tag{$*$}$$

Now, let $n = \frac{1}{x}$.

Then substituting $n$ into $(*)$ and noting that as $x \to 0^{+}$ we have $n \to \infty$, it follows that

$$\lim_{n \to \infty}\left( 1 + \frac{1}{n} \right)^{n} = e.$$
This can be done straight from the definition of $\ln$ as $\ln(x) \equiv \int_{1}^x dt/t$ and $\exp$ as its inverse function. Then we have
\begin{multline}
\ln((1+1/n)^n) = \int_1^{(1+1/n)^n} \frac{dt}{t} = \int_1^{(1+1/n)^{n-1}} \frac{dt}{t}  + \int_{(1+1/n)^{n-1}}^{(1+1/n)^n} \frac{dt}{t} \\ = \int_1^{(1+1/n)^{n-1}} \frac{dt}{t} + \int_1^{1+1/n} \frac{du}{u} = \ln((1+1/n)^{n-1}) +  \int_1^{1+1/n} \frac{du}{u}.
\end{multline}
Repeating this $n$ times gives
$$
\ln((1+1/n)^n) = n\int_1^{1+1/n} \frac{du}{u}.
$$
It's not hard to show from the properties of the integral that
$$
\lim_{n\rightarrow\infty} n\int_1^{1+1/n} \frac{du}{u} = 1,
$$
and that $\ln$ is continuous, so it is valid to say
$$
\lim_{n\rightarrow\infty}\ln((1+1/n)^n) = \ln\left(\lim_{n\rightarrow\infty}(1+1/n)^n.\right)
$$
Thus,
$$
\ln\left(\lim_{n\rightarrow\infty}(1+1/n)^n\right) = 1
$$
and so we have
$$
\lim_{n\rightarrow\infty}(1+1/n)^n = \exp(1) \equiv e
$$