Question

How can I solve this problem without using L'hopital? \[\ln \left( {\ln x} \right)\] as \[x\] goes +infinity

Answer 1

Hint: In this question we are asked to find the limit of the given function without using L’hopital method, for this we have to substitute the limit value directly in the given function and calculate the value then we will get the required answer.

Complete step by step answer:
A limit is defined as a function that has some value that approaches the input. A limit of a function is represented as:
\[\mathop {\lim }\limits_{x \to n} f\left( x \right) = L\] ,
Here lim refers to limit, it generally describes that the real valued function \[f\left( x \right)\] tends to attain the limit L as \[x\] tends to n and is denoted by an arrow.
We can read this as “the limit of any given function ‘f’ of ‘ \[x\] ’ as ‘ \[x\] ’ approaches to \[n\] is equal to L.
Given expression is \[\ln \left( {\ln x} \right)\] ,
Applying limit we get,
\[\mathop {\lim }\limits_{x \to {\infty ^ + }} \ln \left( {\ln x} \right)\] ,
Here as \[x\] increases without bound, \[\ln x\] also increases without bound. Therefore \[\ln \left( {\ln x} \right)\] increases without bound.
So, applying limit directly as we are asked to find the limit without using the L’hopital method, we get,
\[ \Rightarrow \mathop {\lim }\limits_{x \to {\infty ^ + }} \ln \left( {\ln x} \right) = \mathop {\lim }\limits_{x \to {\infty ^ + }} \ln \left( {\ln \infty } \right)\] ,
We know that \[\ln x = \infty \] , now substituting the value we get,
\[ \Rightarrow \mathop {\lim }\limits_{x \to {\infty ^ + }} \ln \left( {\ln x} \right) = \mathop {\lim }\limits_{x \to {\infty ^ + }} \ln \infty \] ,
Again substituting the value \[\ln x = \infty \] in the above expression,
\[ \Rightarrow \mathop {\lim }\limits_{x \to {\infty ^ + }} \ln \left( {\ln x} \right) = \infty \] ,
So the limit value will be equal to \[\infty \] .

\[\therefore \] The value of the limit of the given function \[\ln \left( {\ln x} \right)\] is equal to \[\infty \] .

Note: L'Hôpital's rule can only be applied in the case where direct substitution yields an indeterminate form, meaning \[\dfrac{0}{0}\] or \[\dfrac{{ \pm \infty }}{{ \pm \infty }}\] , So if f and g are defined, L'Hôpital would be applicable only if the value of both f and g is 0. In these types of questions we cannot find the limit without using L’hospital method.