# Question #060c5

May 22, 2016

$1$

#### Explanation:

We can rewrite this as a fraction so that l'Hospital's rule applies:

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

Note that this is in the indeterminate form $\frac{0}{0}$, so take the derivative of the numerator and denominator.

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

Simplifying, this becomes:

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

Here, we see that $\frac{1}{x}$ will go to $0$ as $x$ approaches infinity, so the limit equals

$= \frac{1}{1 + 0} = 1$

Jan 14, 2018

${\lim}_{x \to \infty} x \ln \left(1 + \frac{1}{x}\right) = 1$

#### Explanation:

Alternatively, we can use the following logarithm property:
$b {\log}_{x} \left(a\right) = {\log}_{x} \left({a}^{b}\right)$

This allows us to bring the $x$ as an exponent:
${\lim}_{x \to \infty} \ln \left({\left(1 + \frac{1}{x}\right)}^{x}\right)$

We can recognize the bit inside the ln function as the definition for the number $e$, so this simply evaluates to:
$\ln \left(e\right) = 1$