Question

How do you find the limit of ${\left( {\ln x} \right)^{\dfrac{1}{x}}}$ as x approaches infinity?

Answer 1

Hint : Here in this question, we need to find the limit for the function. Let we define the given function as y and we apply the limit to it. Here we can’t apply the limit directly so by using the concept of log and antilog and then we apply the limit. The function will get closer and closer to some number. When the x approaches to $\infty$ we need to find the limit of a function.

Complete step-by-step answer :
The idea of a limit is a basis of all calculus. The limit of a function is defined as let $f(x)$ be a function defined on an interval that contains $x = a$ . Then we say that $\mathop {\lim }\limits_{x \to a} f(x) = L$ , if for every $\varepsilon > 0$ there is some number $\delta > 0$ such that $\left| {f(x) - L} \right| < \varepsilon$ whenever $0 < \left| {x - a} \right| < \delta$ .
Here we have to find the value of the limit when x tends to infinity. Let we define the given function as $y = {\left( {\ln x} \right)^{\dfrac{1}{x}}}$ .
Put natural log on the both sides we have
$\Rightarrow \ln y = \ln \left( {{{\left( {\ln x} \right)}^{\dfrac{1}{x}}}} \right)$
By using the property of the natural logarithmic we are going to write the above equation as
$\Rightarrow \ln y = \dfrac{1}{x}\ln \left( {\ln x} \right)$
The above equation is rewritten as
$\Rightarrow \ln y = \dfrac{{\ln \left( {\ln x} \right)}}{x}$
Now we are going to apply the limit to the function $y$ so we have
$\Rightarrow \mathop {\lim }\limits_{x \to \infty } \ln y = \mathop {\lim }\limits_{x \to \infty } \dfrac{{\ln (\ln x)}}{x}$
When we apply the limit we obtain
$\Rightarrow \mathop {\lim }\limits_{x \to \infty } \ln y = \dfrac{\infty }{\infty }$
Which is an indeterminate form. By applying L'Hopital's rule the function is written as
$\Rightarrow \mathop {\lim }\limits_{x \to \infty } \ln y = \mathop {\lim }\limits_{x \to \infty } \dfrac{{\dfrac{1}{{\ln x}}.\dfrac{1}{x}}}{1}$
$\Rightarrow \mathop {\lim }\limits_{x \to \infty } \ln y = \mathop {\lim }\limits_{x \to \infty } \dfrac{1}{{x(\ln x)}}$
When we apply the limit we get
$\Rightarrow \ln y = \dfrac{1}{{\infty (\ln \infty )}}$
When any number multiplied by $\infty$ then the answer will be $\infty$ , so we have
$\Rightarrow \ln y = \dfrac{1}{\infty }$
Any number divided by infinity the answer will be zero
$\Rightarrow \ln y = 0$
Apply anti natural log on both sides. The anti-natural log is $e$ . So we have
$\Rightarrow {e^{\ln y}} = {e^0}$
The anti-natural log and natural log will cancel each other and any number raised to the power 0 then it is 1. So we have
$\Rightarrow y = 1$
Therefore we have found the limit for the function $y = 1$
Hence $\mathop {\lim }\limits_{x \to \infty } {(\ln x)^{\dfrac{1}{x}}} = 1$
So, the correct answer is “1”.

Note : When we apply the limit to the function if we get the indeterminate form when we apply the L'Hospital's rule to the function. It is defined as $\mathop {\lim }\limits_{x \to c} \dfrac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to c} \dfrac{{f'(x)}}{{g'(x)}}$ . Then we are going to apply the limit to the function. When the function is a logarithmic function, we apply anti-log to the function. Sometimes when we apply anti log the function will get simplified.