Question

How do you find the limit of $\dfrac{{{{\left( {\ln x} \right)}^2}}}{x}$ as $x$ approaches infinity?

Answer 1

Hint : L’Hospital’s rule provides a technique to evaluate limits of indeterminate form. If in a given limit $\mathop {\lim }\limits_{x \to c} \dfrac{{f(x)}}{{g(x)}}$ , substituting $x$ with $c$ in the function $\dfrac{{f(x)}}{{g(x)}}$ gives an indeterminate form like $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$ , we apply L'Hopital's rule. According to this rule, we have to substitute both the functions in the numerator and denominator with their respective derivatives.
i.e., $\mathop {\lim }\limits_{x \to c} \dfrac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to c} \dfrac{{f'(x)}}{{g'(x)}}$
And then, we again substitute $x$ with $c$ in the function $\dfrac{{f'(x)}}{{g'(x)}}$ to check the value. If the value is a definite number, we have obtained the answer. But if it again yields an indeterminate form, we continue to apply the rule by substituting both the functions in the numerator and denominator with their respective derivatives till we obtain a definite value.
So check whether L’Hospital’s rule is applicable or not. If yes, try to apply and get the answer.

Complete step-by-step answer :
(i)
We are given,
$\mathop {\lim }\limits_{x \to \infty } \dfrac{{{{\left( {\ln x} \right)}^2}}}{x}$
When we substitute $x$ as $\infty$ in $\dfrac{{{{\left( {\ln x} \right)}^2}}}{x}$ , we obtain:
$\dfrac{{{{\left( {\ln \infty } \right)}^2}}}{\infty }$
Since, $\ln \infty = \infty$ , we get:
$\dfrac{{{{\left( \infty \right)}^2}}}{\infty } = \dfrac{\infty }{\infty }$ i.e., an indeterminate form.
(ii)
As we know that L’Hospital’s rule helps to evaluate limits of indeterminate form, we will apply it here. So according to L’Hospital’s rule,
$\mathop {\lim }\limits_{x \to c} \dfrac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to c} \dfrac{{f'(x)}}{{g'(x)}}$
Here, $f(x) = {\left( {\ln x} \right)^2}$ and $g(x) = x$
Since we know that, $\dfrac{d}{{dx}}\left( {\ln x} \right) = \dfrac{1}{x}$
Therefore, by chain rule,
$f'(x) = 2 \times \ln x \times \dfrac{1}{x}$
i.e., $f'(x) = \dfrac{{2\left( {\ln x} \right)}}{x}$
and $g'(x) = 1$
(iii)
Now, applying L’Hospital’s rule i.e., substituting the numerator and denominator with their respective derivatives, we get:
$\mathop {\lim }\limits_{x \to \infty } \dfrac{{{{\left( {\ln x} \right)}^2}}}{x} = \mathop {\lim }\limits_{x \to \infty } \dfrac{{2\left( {\ln x} \right)}}{{x \times 1}}$
i.e.,
$\mathop {\lim }\limits_{x \to \infty } \dfrac{{{{\left( {\ln x} \right)}^2}}}{x} = \mathop {\lim }\limits_{x \to \infty } \dfrac{{2\left( {\ln x} \right)}}{x}$
(iv)
Now, we will again check by substituting $x$ as $\infty$ in $\dfrac{{2\left( {\ln x} \right)}}{x}$ . We will get:
$\dfrac{{2\left( {\ln \infty } \right)}}{\infty }$
Since, $\ln \infty = \infty$ , we get:
$\dfrac{{2\left( \infty \right)}}{\infty } = \dfrac{\infty }{\infty }$ i.e., an indeterminate form.
Therefore, we will again apply the L’Hospital’s rule and will replace the numerator and the denominator with their derivatives respectively.
This time we have $f\left( x \right) = 2\left( {\ln x} \right)$ . Therefore,
$f'\left( x \right) = 2\left( {\dfrac{1}{x}} \right)$
And since, $g\left( x \right) = x$ . Therefore,
$g'\left( x \right) = 1$
(v)
Now, we have:
$\mathop {\lim }\limits_{x \to \infty } \dfrac{{2\left( {\ln x} \right)}}{x} = \mathop {\lim }\limits_{x \to \infty } \dfrac{2}{x}$
Putting $x = \infty$ in $\dfrac{2}{x}$ gives $0$ which is a definite value.
Therefore, $\mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{2}{x}} \right) = 0$
And hence, $\mathop {\lim }\limits_{x \to \infty } \dfrac{{{{\left( {\ln x} \right)}^2}}}{x} = 0$

Note : Note that the graph of $\ln x$ approaches infinity when $x$ is infinity i.e., $\ln \infty = \infty$ . We have to repeat the L’Hospital’s rule till we get a definite value instead of an indeterminate form. Here, we got $\dfrac{1}{\infty }$ which we know is $0$ and not an indeterminate form. So, it will be our final answer. Do not confuse $0$ by $\dfrac{0}{0}$ form as $0$ is a definite value and $\dfrac{0}{0}$ is an indeterminate form.