Proving $\ln e = 1$ Using the definition $$ \ln x = \int_1^x \frac{dt}{t}, $$ is it possible to show that $\ln e = 1$ without showing first that $\exp$ and $\ln$ are inverse functions? Here, $e$ is defined by the series $$ e = \sum_{k=0}^\infty \frac{1}{k!}. $$

EDIT: A useful intermediate step in showing this result is $$ \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^n = e. $$ However, the usual proof of this limit by L'Hospital's rule uses the fact that $\exp$ and $\ln$ are inverse functions. Is there an alternate proof that does not require the inverse property?
Remember, we are working from the series definition of $e$ stated above.
 A: We have $\displaystyle \ln(x)=\int_1^x \frac{dt}{t}$.  Set $\displaystyle g(x)=\sum_n \frac{x^n}{n!}$. Then $g(1)=e$, $g'(x)=g(x)$, and so $\ln(g(x))'=\ln'(g(x))g'(x)=\frac{1}{g(x)}g(x)=1$, hence $\ln(g(x))=x+c$. Since $g(0)=1$ and $\ln(1)=0$, we have $c=0$, hence $\ln(g(x))=x$, and so $\ln(e)=\ln(g(1))=1$.
A: It can be shown quite easily (must add: using ONLY the binomial theorem, grumpy kid) that $$e=\sum_{k=0}^\infty \frac{1}{k!}=\lim_{k\rightarrow\infty}(1+1/k)^k$$
If you don't want to get too picky about mathematical rigor, you will accept the following steps
$$\ln (\lim_{k\rightarrow\infty}(1+1/k)^k)=\lim_{k\rightarrow\infty}\ln ((1+1/k)^k)$$
$$ =\lim_{k\rightarrow\infty}k\ln ((1+1/k))$$
$$ =\lim_{k\rightarrow\infty}\frac{\ln ((1+1/k))}{1/k}$$
$$ =\lim_{u\rightarrow 0}\frac{\ln (1+u)}{u}$$
Applying L'hospital and the fundamental theorem of calculus
$$ =\lim_{u\rightarrow 0}\frac{1}{u+1}=1$$
A: By the definition, we have
$$
e = \lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n.
$$
Now, applying $\ln$ both sides.
$$
\begin{align}
\ln e &= \lim_{n \to \infty}\ln\left(1 + \frac{1}{n}\right)^n\\
\ln e&=\lim_{n \to \infty}n\ln\left(1 + \frac{1}{n}\right).
\end{align}
$$
Let $\dfrac{1}{n}=x\;\Rightarrow\; n=\dfrac{1}{x}$. As $n\to\infty$, $x\to0$, then
$$
\ln e=\lim_{x \to 0}\frac{\ln\left(1 + x\right)}{x}.
$$
Now you can apply L'Hospital's rule to the RHS. I hope this helps.
$$\\$$

$$\Large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$
A: One can show that $e = \lim_{n \to \infty} \left( 1 + \frac 1n\right)^n$ using only the binomial theorem. 


*

*Consider $a_n = (1 + \frac 1n )^n$.

*By the binomial theorem, 
$$\begin{align}
a_n &= \sum_{0 \leq k \leq n} {n \choose k} \left( \frac 1n \right)^k \\
&= 1 + 1 + \frac{1}{2!}(1 - \frac 1n) + \frac{1}{3!}(1 - \frac 1n)(1 - \frac 2n) + \ldots + \frac{1}{n!}(1 - \frac 1n)\cdots(1 - \frac{n-1}{n}).
\end{align}$$
From this, it's quite easy to see that $a_n < a_{n+1}$ and $a_n < e$ using your definition of $e$, so this limit exists.

*Fix some $m$ and consider the terms just up to $nm$:
$$ 1 + 1 + \frac{1}{2!}(1 - \frac 1n) + \dots + \frac{1}{m!}(1 - \frac 1n)\cdots(1 - \frac{m-1}{n}),$$
which is clearly less than $a_n$, and as $n \to \infty$, this goes to
$$ 1 + \frac{1}{1!} + \frac{1}{2!} + \dots + \frac{1}{m!}.$$
As we can do this for any $m < n$, taking the limits as $n \to \infty$ and then $m \to \infty$ gives us that $e \leq \lim a_n \leq e$, or that $e = \lim a_n$, which is what I sought to prove. 


Now we see this doesn't use that exponentials and logarithms are inverses at all, and completes the proof.
A: We know that $e= \lim_{n \to +\infty}(1+\frac 1n)^{n}$.
Consider $\frac d{dx}(ln x)=\lim_{h \to0}\frac {ln(x+h)-ln(x)}h\ =\ lim_{h \to 0}\frac {ln(\frac {x+h}x)}h$ = $\lim_{h \to0}\frac 1hln(1+\frac hx)$
Let $u=\frac xh$ hence as $h\to 0$, $u\to \infty$. Thus we have
$\frac d{dx}(ln x)=\lim_{u \to +\infty}(\frac ux(ln(1+\frac 1u)))$ = $\lim_{u \to +\infty} (\frac 1x(ln(1+\frac 1u)^{u})$ = $\frac 1x lne$.
Now integrate both sides to get the result.
