# Evaluate the limit lim_(x rarr 0^+) (1+1/x)^x ?

Jul 23, 2017

${\lim}_{x \rightarrow {0}^{+}} {\left(1 + \frac{1}{x}\right)}^{x} = 1$

#### Explanation:

${\lim}_{x \rightarrow {0}^{+}} \frac{1}{x} = 0$, so

${\lim}_{x \rightarrow {0}^{+}} \left(1 + \frac{1}{x}\right) = 1$, and

${\lim}_{x \rightarrow {0}^{+}} {\left(1 + \frac{1}{x}\right)}^{x} = {1}^{0} = 1$

Jul 23, 2017

${\lim}_{x \rightarrow {0}^{+}} {\left(1 + \frac{1}{x}\right)}^{x} = 1$

#### Explanation:

We seek:

$L = {\lim}_{x \rightarrow {0}^{+}} {\left(1 + \frac{1}{x}\right)}^{x}$

As the log function is monotonic we can tak logs of both sides to get:

$\ln L = \ln \left\{{\lim}_{x \rightarrow {0}^{+}} {\left(1 + \frac{1}{x}\right)}^{x}\right\}$
$\setminus \setminus \setminus \setminus \setminus \setminus = {\lim}_{x \rightarrow {0}^{+}} \ln \left\{{\left(1 + \frac{1}{x}\right)}^{x}\right\}$
$\setminus \setminus \setminus \setminus \setminus \setminus = {\lim}_{x \rightarrow {0}^{+}} x \ln \left(1 + \frac{1}{x}\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus = {\lim}_{x \rightarrow {0}^{+}} x \ln \left(\frac{x + 1}{x}\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus = {\lim}_{x \rightarrow {0}^{+}} x \left\{\ln \left(x + 1\right) - \ln x\right\}$
$\setminus \setminus \setminus \setminus \setminus \setminus = {\lim}_{x \rightarrow {0}^{+}} \left\{x \ln \left(x + 1\right) - x \ln x\right\}$
$\setminus \setminus \setminus \setminus \setminus \setminus = {\lim}_{x \rightarrow {0}^{+}} x \ln \left(x + 1\right) - {\lim}_{x \rightarrow {0}^{+}} x \ln x$
$\setminus \setminus \setminus \setminus \setminus \setminus = 0 - 0$

Thus:

$L = {e}^{0} = 1$