# Find the limit. Use L'Hospital's Rule where appropriate. If there is a more elementary method,...

## Question:

Find the limit. Use L'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If L'Hospital's Rule doesn't apply, explain why.

{eq}\lim_{x \to \infty}\frac{\textrm{ln} \; \sqrt{x}}{x^2} {/eq}

## L'Hopital's Rule:

To apply L'Hopital's Rule we must have an indeterminacy of the form {eq}\dfrac{0}{0} {/eq} and {eq}\dfrac{ \pm \infty }{ \pm \infty }. {/eq}

L'Hopital's rule consists of deriving the numerator and denominator separately.

The application of the rule does not imply the existence of the limit.

## Answer and Explanation: 1

Let's find the limit.

{eq}\displaystyle \lim_{x \to \infty}\frac{\textrm{ln} \; \sqrt{x}}{x^2} {/eq}.

Evaluating we have

{eq}\begin{align*} \displaystyle \lim_{x \to \infty} \dfrac{\textrm{ln} \; \sqrt{x}}{x^2} & = \dfrac{\textrm{ln} \; \sqrt{ \infty }}{\infty ^2} \\ &= \dfrac{\infty }{ \infty } \end{align*} {/eq}.

As {eq}\displaystyle \lim_{x \to \infty} \dfrac{\textrm{ln} \; \sqrt{x}}{x^2} = \dfrac{\infty }{ \infty } {/eq}, we can use the L'Hopital Rule.

But before applying L'Hopital let's redefine the limit by applying the properties of the natural logarithm.

Then

{eq}\displaystyle \lim_{x \to \infty} \dfrac{\textrm{ln} \; \sqrt{x}}{x^2} = \displaystyle \lim_{x \to \infty} \dfrac{1}{2} \dfrac{\textrm{ln}\; x}{x^2}. {/eq}

Applying L'Hopital, we have

{eq}\begin{align*} \displaystyle \lim_{x \to \infty} \dfrac{1}{2} \dfrac{\textrm{ln}\; x}{x^2} &= \dfrac{1}{2}\displaystyle \lim_{x \to \infty} \dfrac{\dfrac{1}{x}}{2x} \\ &= \dfrac{1}{2}\displaystyle \lim_{x \to \infty} \dfrac{1}{2x^2}\\ &= \dfrac{1}{\infty } \\ &=0. \end{align*} {/eq}

Therefore,

{eq}\boxed{\displaystyle \lim_{x \to \infty} \dfrac{\textrm{ln} \; \sqrt{x}}{x^2} =0}. {/eq}

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Chapter 9 / Lesson 9Explore L'Hopital's rule and learn who came up with this rule. See examples to understand L'Hopital's rule when limit approaches zero and when limit approaches infinity.