# 4.8 L’hôpital’s rule  (Page 3/7)

 Page 3 / 7

Evaluate $\underset{x\to \infty }{\text{lim}}\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}x}{5x}.$

$0$

As mentioned, L’Hôpital’s rule is an extremely useful tool for evaluating limits. It is important to remember, however, that to apply L’Hôpital’s rule to a quotient $\frac{f\left(x\right)}{g\left(x\right)},$ it is essential that the limit of $\frac{f\left(x\right)}{g\left(x\right)}$ be of the form $\frac{0}{0}$ or $\infty \text{/}\infty .$ Consider the following example.

## When l’hôpital’s rule does not apply

Consider $\underset{x\to 1}{\text{lim}}\frac{{x}^{2}+5}{3x+4}.$ Show that the limit cannot be evaluated by applying L’Hôpital’s rule.

Because the limits of the numerator and denominator are not both zero and are not both infinite, we cannot apply L’Hôpital’s rule. If we try to do so, we get

$\frac{d}{dx}\left({x}^{2}+5\right)=2x$

and

$\frac{d}{dx}\left(3x+4\right)=3.$

At which point we would conclude erroneously that

$\underset{x\to 1}{\text{lim}}\frac{{x}^{2}+5}{3x+4}=\underset{x\to 1}{\text{lim}}\frac{2x}{3}=\frac{2}{3}.$

However, since $\underset{x\to 1}{\text{lim}}\left({x}^{2}+5\right)=6$ and $\underset{x\to 1}{\text{lim}}\left(3x+4\right)=7,$ we actually have

$\underset{x\to 1}{\text{lim}}\frac{{x}^{2}+5}{3x+4}=\frac{6}{7}.$

We can conclude that

$\underset{x\to 1}{\text{lim}}\frac{{x}^{2}+5}{3x+4}\ne \underset{x\to 1}{\text{lim}}\frac{\frac{d}{dx}\left({x}^{2}+5\right)}{\frac{d}{dx}\left(3x+4\right)}.$

Explain why we cannot apply L’Hôpital’s rule to evaluate $\underset{x\to {0}^{+}}{\text{lim}}\frac{\text{cos}\phantom{\rule{0.1em}{0ex}}x}{x}.$ Evaluate $\underset{x\to {0}^{+}}{\text{lim}}\frac{\text{cos}\phantom{\rule{0.1em}{0ex}}x}{x}$ by other means.

$\underset{x\to {0}^{+}}{\text{lim}}\text{cos}\phantom{\rule{0.1em}{0ex}}x=1.$ Therefore, we cannot apply L’Hôpital’s rule. The limit of the quotient is $\infty$

## Other indeterminate forms

L’Hôpital’s rule is very useful for evaluating limits involving the indeterminate forms $\frac{0}{0}$ and $\infty \text{/}\infty .$ However, we can also use L’Hôpital’s rule to help evaluate limits involving other indeterminate forms that arise when evaluating limits. The expressions $0·\infty ,$ $\infty -\infty ,$ ${1}^{\infty },$ ${\infty }^{0},$ and ${0}^{0}$ are all considered indeterminate forms. These expressions are not real numbers. Rather, they represent forms that arise when trying to evaluate certain limits. Next we realize why these are indeterminate forms and then understand how to use L’Hôpital’s rule in these cases. The key idea is that we must rewrite the indeterminate forms in such a way that we arrive at the indeterminate form $\frac{0}{0}$ or $\infty \text{/}\infty .$

## Indeterminate form of type $0·\infty$

Suppose we want to evaluate $\underset{x\to a}{\text{lim}}\left(f\left(x\right)·g\left(x\right)\right),$ where $f\left(x\right)\to 0$ and $g\left(x\right)\to \infty$ (or $\text{−}\infty \right)$ as $x\to a.$ Since one term in the product is approaching zero but the other term is becoming arbitrarily large (in magnitude), anything can happen to the product. We use the notation $0·\infty$ to denote the form that arises in this situation. The expression $0·\infty$ is considered indeterminate because we cannot determine without further analysis the exact behavior of the product $f\left(x\right)g\left(x\right)$ as $x\to \infty .$ For example, let $n$ be a positive integer and consider

$f\left(x\right)=\frac{1}{\left({x}^{n}+1\right)}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}g\left(x\right)=3{x}^{2}.$

As $x\to \infty ,$ $f\left(x\right)\to 0$ and $g\left(x\right)\to \infty .$ However, the limit as $x\to \infty$ of $f\left(x\right)g\left(x\right)=\frac{3{x}^{2}}{\left({x}^{n}+1\right)}$ varies, depending on $n.$ If $n=2,$ then $\underset{x\to \infty }{\text{lim}}f\left(x\right)g\left(x\right)=3.$ If $n=1,$ then $\underset{x\to \infty }{\text{lim}}f\left(x\right)g\left(x\right)=\infty .$ If $n=3,$ then $\underset{x\to \infty }{\text{lim}}f\left(x\right)g\left(x\right)=0.$ Here we consider another limit involving the indeterminate form $0·\infty$ and show how to rewrite the function as a quotient to use L’Hôpital’s rule.

## Indeterminate form of type $0·\infty$

Evaluate $\underset{x\to {0}^{+}}{\text{lim}}x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}x.$

First, rewrite the function $x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}x$ as a quotient to apply L’Hôpital’s rule. If we write

$x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}x=\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}x}{1\text{/}x},$

we see that $\text{ln}\phantom{\rule{0.1em}{0ex}}x\to \text{−}\infty$ as $x\to {0}^{+}$ and $\frac{1}{x}\to \infty$ as $x\to {0}^{+}.$ Therefore, we can apply L’Hôpital’s rule and obtain

$\underset{x\to {0}^{+}}{\text{lim}}\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}x}{1\text{/}x}=\underset{x\to {0}^{+}}{\text{lim}}\frac{\frac{d}{dx}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}x\right)}{\frac{d}{dx}\left(1\text{/}x\right)}=\underset{x\to {0}^{+}}{\text{lim}}\frac{1\text{/}{x}^{2}}{-1\text{/}x}=\underset{x\to {0}^{+}}{\text{lim}}\left(\text{−}x\right)=0.$

We conclude that

$\underset{x\to {0}^{+}}{\text{lim}}x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}x=0.$ Finding the limit at x = 0 of the function f ( x ) = x ln x .

Evaluate $\underset{x\to 0}{\text{lim}}x\phantom{\rule{0.1em}{0ex}}\text{cot}\phantom{\rule{0.1em}{0ex}}x.$

$1$

## Indeterminate form of type $\infty -\infty$

Another type of indeterminate form is $\infty -\infty .$ Consider the following example. Let $n$ be a positive integer and let $f\left(x\right)=3{x}^{n}$ and $g\left(x\right)=3{x}^{2}+5.$ As $x\to \infty ,$ $f\left(x\right)\to \infty$ and $g\left(x\right)\to \infty .$ We are interested in $\underset{x\to \infty }{\text{lim}}\left(f\left(x\right)-g\left(x\right)\right).$ Depending on whether $f\left(x\right)$ grows faster, $g\left(x\right)$ grows faster, or they grow at the same rate, as we see next, anything can happen in this limit. Since $f\left(x\right)\to \infty$ and $g\left(x\right)\to \infty ,$ we write $\infty -\infty$ to denote the form of this limit. As with our other indeterminate forms, $\infty -\infty$ has no meaning on its own and we must do more analysis to determine the value of the limit. For example, suppose the exponent $n$ in the function $f\left(x\right)=3{x}^{n}$ is $n=3,$ then

determine the area of the region enclosed by x²+y=1,2x-y+4=0
How to use it to slove fraction
Hello please can someone tell me the meaning of this group all about, yes I know is calculus group but yet nothing is showing up
Shodipo
You have downloaded the aplication Calculus Volume 1, tackling about lessons for (mostly) college freshmen, Calculus 1: Differential, and this group I think aims to let concerns and questions from students who want to clarify something about the subject. Well, this is what I guess so.
Jean
Im not in college but this will still help
nothing
how can we scatch a parabola graph
Ok
Endalkachew
how can I solve differentiation?
with the help of different formulas and Rules. we use formulas according to given condition or according to questions
CALCULUS
For example any questions...
CALCULUS
what is the procedures in solving number 1?
review of funtion role?
for the function f(x)={x^2-7x+104 x<=7 7x+55 x>7' does limx7 f(x) exist?
find dy÷dx (y^2+2 sec)^2=4(x+1)^2
Integral of e^x/(1+e^2x)tan^-1 (e^x)
why might we use the shell method instead of slicing
fg[[(45)]]²+45⅓x²=100
find the values of c such that the graph of f(x)=x^4+2x^3+cx^2+2x+2
anyone to explain some basic in calculus
I can
Debdoot
A conical container of radius 10 ft and height 30 ft is filled with water to a depth of 15 ft. How much work is required to pump all the water out through a hole in the top of the container if the unit weight of the water is 62.4 lb/ft^3?
hi am new here I really wants to know how the solve calculus
IBRAHIM
me too. I want to know calculation involved in calculus.
Katiba
evaluate triple integral xyz dx dy dz where the domain v is bounded by the plane x+y+z=a and the co-ordinate planes
So how can this question be solved
Eddy
i m not sure but it could be xyz/2
Leo
someone should explain with a photo shot of the working pls
I think we should sort it out.
Eunice
Eunice Toe you can try it if you have the idea
how
Eunice
a^6÷8
Muzamil
i think a^6 ÷ 8
Muzamil By OpenStax By Laurence Bailen By Robert Morris By Robert Morris By Angela January By OpenStax By Brenna Fike By Angela Eckman By Zarina Chocolate By Sheila Lopez