# Nice question in calculas?

## prove that $\ln e = 1$

Apr 28, 2018

I am nit sure it is a proof....but anyway:

#### Explanation:

From the definition of log:

${\log}_{b} x = a$

auch that:

$x = {b}^{a}$

that is, the integrand is equal to the base to the power of the result.

In our case we know that the Natural Logarithm, $\ln$, is the one with base $e$, i.e.:

${\log}_{e} = \ln$

so, using our definition of log:

$\ln e = 1$
${\log}_{e} e = 1$
so that:
$e = {e}^{1}$ that is correct.

Apr 28, 2018

Show the steps below

#### Explanation:

show the explanation

$\ln e = \ln \left[{\lim}_{x \rightarrow \infty} {\left(1 + \frac{1}{x}\right)}^{x}\right]$

$= \left[{\lim}_{x \rightarrow \infty} \ln {\left(1 + \frac{1}{x}\right)}^{x}\right]$

$= \left[{\lim}_{x \rightarrow \infty} x \cdot \ln \left(1 + \frac{1}{x}\right)\right]$ product $\left(0 \cdot \infty\right)$

$= \left[{\lim}_{x \rightarrow \infty} \frac{\ln \left(1 + \frac{1}{x}\right)}{\frac{1}{x}}\right]$ product $\left(\frac{0}{0}\right)$

$= \left[{\lim}_{x \rightarrow \infty} \frac{- \frac{\frac{1}{x} ^ 2}{1 + \frac{1}{x}}}{- \frac{1}{x} ^ 2}\right] = 1$

Apr 28, 2018

Nearly all such proofs are cyclic.

Using Bernoulli's definition , we have:

$e = {\lim}_{n \rightarrow \infty} {\left(1 + \frac{1}{n}\right)}^{n}$

If we apply base $e$ (Napier) logarithms to both sides:

$\ln e = \ln \setminus {\lim}_{n \rightarrow \infty} {\left(1 + \frac{1}{n}\right)}^{n}$

Due to the monotonicity of the logarithm function:

$\ln e = {\lim}_{n \rightarrow \infty} \ln {\left(1 + \frac{1}{n}\right)}^{n}$

Using the rules of logarithms:

$\ln e = {\lim}_{n \rightarrow \infty} n \ln \left(1 + \frac{1}{n}\right)$

We can simplify by performing a substitution:

$u = \frac{1}{n} \iff n = \frac{1}{u}$, and as $n \rightarrow \infty \implies u \rightarrow 0$

Thus:

$\ln e = {\lim}_{u \rightarrow 0} \frac{1}{u} \ln \left(1 + u\right)$

$\setminus \setminus \setminus \setminus \setminus = {\lim}_{u \rightarrow 0} \frac{\ln \left(1 + u\right)}{u}$

This is of an indeterminate for $0 / 0$, so we can apply L'Hôpital's rule, to get:

$\ln e = {\lim}_{u \rightarrow 0} \frac{\frac{d}{\mathrm{du}} \ln \left(1 + u\right)}{\frac{d}{\mathrm{du}} u}$

$\setminus \setminus \setminus \setminus \setminus = {\lim}_{u \rightarrow 0} \frac{\frac{1}{1 + u}}{1}$

$\setminus \setminus \setminus \setminus \setminus = 1 \setminus \setminus \setminus$ QED