1. Limit: Indeterminate Form/L'Hospital's Rule

Evaluate the limit. (If you need to use - or , enter -INFINITY or INFINITY.)

$
\mathop {\lim }\limits_{x \to \infty } \frac{{\ln \left( {\ln \left( {4x} \right)} \right)}}
{{4x}}
$

Here's what I tried. Here's our first rule:

$
\begin{gathered}
\mathop {\lim }\limits_{x \to \infty } \frac{{f(x)}}
{{g(x)}} = \mathop {\lim }\limits_{x \to \infty } \frac{{f'(x)}}
{{g'(x)}} \hfill \\
\mathop {\lim }\limits_{x \to \infty } \frac{{\tfrac{1}
{{\ln 4x}} \cdot \tfrac{1}
{{4x}} \cdot \tfrac{4}
{1}}}
{4} \hfill \\
\mathop {\lim }\limits_{x \to \infty } \frac{{\tfrac{1}
{{x\ln 4x}}}}
{4} \hfill \\
\end{gathered}
$

$
\begin{gathered}
\mathop {\lim }\limits_{x \to \infty } \frac{{\tfrac{1}
{{x\ln 4x}}}}
{4} \hfill \\
\mathop {\lim }\limits_{x \to \infty } \frac{1}
{{4x\ln 4x}} \hfill \\
\end{gathered}
$

So then what? It's not infinity or -infinity.

2. Originally Posted by RedBarchetta
Evaluate the limit. (If you need to use - or , enter -INFINITY or INFINITY.)

$
\mathop {\lim }\limits_{x \to \infty } \frac{{\ln \left( {\ln \left( {4x} \right)} \right)}}
{{4x}}
$

Here's what I tried. Here's our first rule:

$
\begin{gathered}
\mathop {\lim }\limits_{x \to \infty } \frac{{f(x)}}
{{g(x)}} = \mathop {\lim }\limits_{x \to \infty } \frac{{f'(x)}}
{{g'(x)}} \hfill \\
\mathop {\lim }\limits_{x \to \infty } \frac{{\tfrac{1}
{{\ln 4x}} \cdot \tfrac{1}
{{4x}} \cdot \tfrac{4}
{1}}}
{4} \hfill \\
\mathop {\lim }\limits_{x \to \infty } \frac{{\tfrac{1}
{{x\ln 4x}}}}
{4} \hfill \\
\end{gathered}
$

$
\begin{gathered}
\mathop {\lim }\limits_{x \to \infty } \frac{{\tfrac{1}
{{x\ln 4x}}}}
{4} \hfill \\
\mathop {\lim }\limits_{x \to \infty } \frac{1}
{{4x\ln 4x}} \hfill \\
\end{gathered}
$

So then what? It's not infinity or -infinity.
$\lim_{x\to{\infty}}\frac{\ln(\ln 4x)}{4x}$

$\lim_{x\to{\infty}}\frac{\frac{1}{x\ln 4x}}{4}$

$\lim_{x\to{\infty}}\frac{1}{4x\ln 4x}=\color{red}\boxed{0}$. This the case because the denominator approaches infinity : $\frac{1}{\infty}\rightarrow 0$.