# Thread: Limit: Indeterminate Form/L'Hospital's Rule

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

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

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

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

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

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

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.)

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

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

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

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

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

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

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