4.8 L’hôpital’s rule  (Page 2/7)

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Proof

We provide a proof of this theorem in the special case when $f,g,{f}^{\prime },$ and ${g}^{\prime }$ are all continuous over an open interval containing $a.$ In that case, since $\underset{x\to a}{\text{lim}}f\left(x\right)=0=\underset{x\to a}{\text{lim}}g\left(x\right)$ and $f$ and $g$ are continuous at $a,$ it follows that $f\left(a\right)=0=g\left(a\right).$ Therefore,

$\begin{array}{ccccc}\hfill \underset{x\to a}{\text{lim}}\frac{f\left(x\right)}{g\left(x\right)}& =\underset{x\to a}{\text{lim}}\frac{f\left(x\right)-f\left(a\right)}{g\left(x\right)-g\left(a\right)}\hfill & & & \text{since}\phantom{\rule{0.2em}{0ex}}f\left(a\right)=0=g\left(a\right)\hfill \\ & =\underset{x\to a}{\text{lim}}\frac{\frac{f\left(x\right)-f\left(a\right)}{x-a}}{\frac{g\left(x\right)-g\left(a\right)}{x-a}}\hfill & & & \text{algebra}\hfill \\ & =\frac{\underset{x\to a}{\text{lim}}\frac{f\left(x\right)-f\left(a\right)}{x-a}}{\underset{x\to a}{\text{lim}}\frac{g\left(x\right)-g\left(a\right)}{x-a}}\hfill & & & \text{limit of a quotient}\hfill \\ & =\frac{{f}^{\prime }\left(a\right)}{{g}^{\prime }\left(a\right)}\hfill & & & \text{definition of the derivative}\hfill \\ & =\frac{\underset{x\to a}{\text{lim}}{f}^{\prime }\left(x\right)}{\underset{x\to a}{\text{lim}}{g}^{\prime }\left(x\right)}\hfill & & & \text{continuity of}\phantom{\rule{0.2em}{0ex}}{f}^{\prime }\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{g}^{\prime }\hfill \\ & =\underset{x\to a}{\text{lim}}\frac{{f}^{\prime }\left(x\right)}{{g}^{\prime }\left(x\right)}.\hfill & & & \text{limit of a quotient}\hfill \end{array}$

Note that L’Hôpital’s rule states we can calculate the limit of a quotient $\frac{f}{g}$ by considering the limit of the quotient of the derivatives $\frac{{f}^{\prime }}{{g}^{\prime }}.$ It is important to realize that we are not calculating the derivative of the quotient $\frac{f}{g}.$

Applying l’hôpital’s rule (0/0 case)

Evaluate each of the following limits by applying L’Hôpital’s rule.

1. $\underset{x\to 0}{\text{lim}}\frac{1-\text{cos}\phantom{\rule{0.1em}{0ex}}x}{x}$
2. $\underset{x\to 1}{\text{lim}}\frac{\text{sin}\left(\pi x\right)}{\text{ln}\phantom{\rule{0.1em}{0ex}}x}$
3. $\underset{x\to \infty }{\text{lim}}\frac{{e}^{1\text{/}x}-1}{1\text{/}x}$
4. $\underset{x\to 0}{\text{lim}}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x-x}{{x}^{2}}$
1. Since the numerator $1-\text{cos}\phantom{\rule{0.1em}{0ex}}x\to 0$ and the denominator $x\to 0,$ we can apply L’Hôpital’s rule to evaluate this limit. We have
$\begin{array}{cc}\hfill \underset{x\to 0}{\text{lim}}\frac{1-\text{cos}\phantom{\rule{0.1em}{0ex}}x}{x}& =\underset{x\to 0}{\text{lim}}\frac{\frac{d}{dx}\left(1-\text{cos}\phantom{\rule{0.1em}{0ex}}x\right)}{\frac{d}{dx}\left(x\right)}\hfill \\ & =\underset{x\to 0}{\text{lim}}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x}{1}\hfill \\ & =\frac{\underset{x\to 0}{\text{lim}}\left(\text{sin}\phantom{\rule{0.1em}{0ex}}x\right)}{\underset{x\to 0}{\text{lim}}\left(1\right)}\hfill \\ & =\frac{0}{1}=0.\hfill \end{array}$
2. As $x\to 1,$ the numerator $\text{sin}\left(\pi x\right)\to 0$ and the denominator $\text{ln}\left(x\right)\to 0.$ Therefore, we can apply L’Hôpital’s rule. We obtain
$\begin{array}{cc}\hfill \underset{x\to 1}{\text{lim}}\frac{\text{sin}\left(\pi x\right)}{\text{ln}\phantom{\rule{0.1em}{0ex}}x}& =\underset{x\to 1}{\text{lim}}\frac{\pi \phantom{\rule{0.1em}{0ex}}\text{cos}\left(\pi x\right)}{1\text{/}x}\hfill \\ & =\underset{x\to 1}{\text{lim}}\left(\pi x\right)\text{cos}\left(\pi x\right)\hfill \\ & =\left(\pi ·1\right)\left(-1\right)=\text{−}\pi .\hfill \end{array}$
3. As $x\to \infty ,$ the numerator ${e}^{1\text{/}x}-1\to 0$ and the denominator $\left(\frac{1}{x}\right)\to 0.$ Therefore, we can apply L’Hôpital’s rule. We obtain
$\underset{x\to \infty }{\text{lim}}\frac{{e}^{1\text{/}x}-1}{\frac{1}{x}}=\underset{x\to \infty }{\text{lim}}\frac{{e}^{1\text{/}x}\left(\frac{-1}{{x}^{2}}\right)}{\left(\frac{-1}{{x}^{2}}\right)}=\underset{x\to \infty }{\text{lim}}{e}^{1\text{/}x}={e}^{0}=1·\underset{x\to \infty }{\text{lim}}\frac{{e}^{1\text{/}x}-1}{\text{ln}\phantom{\rule{0.1em}{0ex}}x}.$
4. As $x\to 0,$ both the numerator and denominator approach zero. Therefore, we can apply L’Hôpital’s rule. We obtain
$\underset{x\to 0}{\text{lim}}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x-x}{{x}^{2}}=\underset{x\to 0}{\text{lim}}\frac{\text{cos}\phantom{\rule{0.1em}{0ex}}x-1}{2x}.$

Since the numerator and denominator of this new quotient both approach zero as $x\to 0,$ we apply L’Hôpital’s rule again. In doing so, we see that
$\underset{x\to 0}{\text{lim}}\frac{\text{cos}\phantom{\rule{0.1em}{0ex}}x-1}{2x}=\underset{x\to 0}{\text{lim}}\frac{\text{−}\text{sin}\phantom{\rule{0.1em}{0ex}}x}{2}=0.$

Therefore, we conclude that
$\underset{x\to 0}{\text{lim}}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x-x}{{x}^{2}}=0.$

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

$1$

We can also use L’Hôpital’s rule to evaluate limits of quotients $\frac{f\left(x\right)}{g\left(x\right)}$ in which $f\left(x\right)\to \text{±}\infty$ and $g\left(x\right)\to \text{±}\infty .$ Limits of this form are classified as indeterminate forms of type $\infty \text{/}\infty .$ Again, note that we are not actually dividing $\infty$ by $\infty .$ Since $\infty$ is not a real number, that is impossible; rather, $\infty \text{/}\infty .$ is used to represent a quotient of limits, each of which is $\infty$ or $\text{−}\infty .$

L’hôpital’s rule $\left(\infty \text{/}\infty$ Case)

Suppose $f$ and $g$ are differentiable functions over an open interval containing $a,$ except possibly at $a.$ Suppose $\underset{x\to a}{\text{lim}}f\left(x\right)=\infty$ (or $\text{−}\infty \right)$ and $\underset{x\to a}{\text{lim}}g\left(x\right)=\infty$ (or $\text{−}\infty \right).$ Then,

$\underset{x\to a}{\text{lim}}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{\text{lim}}\frac{{f}^{\prime }\left(x\right)}{{g}^{\prime }\left(x\right)},$

assuming the limit on the right exists or is $\infty$ or $\text{−}\infty .$ This result also holds if the limit is infinite, if $a=\infty$ or $\text{−}\infty ,$ or the limit is one-sided.

Applying l’hôpital’s rule $\left(\infty \text{/}\infty$ Case)

Evaluate each of the following limits by applying L’Hôpital’s rule.

1. $\underset{x\to \infty }{\text{lim}}\frac{3x+5}{2x+1}$
2. $\underset{x\to {0}^{+}}{\text{lim}}\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}x}{\text{cot}\phantom{\rule{0.1em}{0ex}}x}$
1. Since $3x+5$ and $2x+1$ are first-degree polynomials with positive leading coefficients, $\underset{x\to \infty }{\text{lim}}\left(3x+5\right)=\infty$ and $\underset{x\to \infty }{\text{lim}}\left(2x+1\right)=\infty .$ Therefore, we apply L’Hôpital’s rule and obtain
$\underset{x\to \infty }{\text{lim}}\frac{3x+5}{2x+1}=\underset{x\to \infty }{\text{lim}}\frac{3+5\text{/}x}{2x+1}=\underset{x\to \infty }{\text{lim}}\frac{3}{2}=\frac{3}{2}.$

Note that this limit can also be calculated without invoking L’Hôpital’s rule. Earlier in the chapter we showed how to evaluate such a limit by dividing the numerator and denominator by the highest power of $x$ in the denominator. In doing so, we saw that
$\underset{x\to \infty }{\text{lim}}\frac{3x+5}{2x+1}=\underset{x\to \infty }{\text{lim}}\frac{3+5\text{/}x}{2x+1}=\frac{3}{2}.$

L’Hôpital’s rule provides us with an alternative means of evaluating this type of limit.
2. Here, $\underset{x\to {0}^{+}}{\text{lim}}\text{ln}\phantom{\rule{0.1em}{0ex}}x=\text{−}\infty$ and $\underset{x\to {0}^{+}}{\text{lim}}\text{cot}\phantom{\rule{0.1em}{0ex}}x=\infty .$ Therefore, we can apply L’Hôpital’s rule and obtain
$\underset{x\to {0}^{+}}{\text{lim}}\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}x}{\text{cot}\phantom{\rule{0.1em}{0ex}}x}=\underset{x\to {0}^{+}}{\text{lim}}\frac{1\text{/}x}{\text{−}{\text{csc}}^{2}x}=\underset{x\to {0}^{+}}{\text{lim}}\frac{1}{\text{−}x\phantom{\rule{0.1em}{0ex}}{\text{csc}}^{2}x}.$

Now as $x\to {0}^{+},$ ${\text{csc}}^{2}x\to \infty .$ Therefore, the first term in the denominator is approaching zero and the second term is getting really large. In such a case, anything can happen with the product. Therefore, we cannot make any conclusion yet. To evaluate the limit, we use the definition of $\text{csc}\phantom{\rule{0.1em}{0ex}}x$ to write
$\underset{x\to {0}^{+}}{\text{lim}}\frac{1}{\text{−}x\phantom{\rule{0.1em}{0ex}}{\text{csc}}^{2}x}=\underset{x\to {0}^{+}}{\text{lim}}\frac{{\text{sin}}^{2}x}{\text{−}x}.$

Now $\underset{x\to {0}^{+}}{\text{lim}}{\text{sin}}^{2}x=0$ and $\underset{x\to {0}^{+}}{\text{lim}}x=0,$ so we apply L’Hôpital’s rule again. We find
$\underset{x\to {0}^{+}}{\text{lim}}\frac{{\text{sin}}^{2}x}{\text{−}x}=\underset{x\to {0}^{+}}{\text{lim}}\frac{2\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}x}{-1}=\frac{0}{-1}=0.$

We conclude that
$\underset{x\to {0}^{+}}{\text{lim}}\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}x}{\text{cot}\phantom{\rule{0.1em}{0ex}}x}=0.$

find the domain and range of f(x)= 4x-7/x²-6x+8
find the range of f(x)=(x+1)(x+4)
-1, -4
Marcia
That's domain. The range is [-9/4,+infinity)
Jacob
If you're using calculus to find the range, you have to find the extrema through the first derivative test and then substitute the x-value for the extrema back into the original equation.
Jacob
Good morning,,, how are you
d/dx{1/y - lny + X^3.Y^5}
How to identify domain and range
hello
Akpevwe
He,,
Harrieta
hi
Dr
hello
velocity
I only talk to girls
Dr
women are smart then guys
Dr
Smarter
sorry
Dr
Dr
:(
Shun
was up
Dr
hello
is it chatting app?.. I do not see any calculus here. lol
Find the arc length of the graph of f(x) = In (sinx) on the interval [Π/4, Π/2].
Sand falling freely from a lorry form a conical shape whose height is always equal to one-third the radius of the base. a. How fast is the volume increasing when the radius of the base is (1m) and increasing at the rate of 1/4cm/sec Pls help me solve
show that lim f(x) + lim g(x)=m+l
list the basic elementary differentials
Differentiation and integration
yes
Damien
proper definition of derivative
the maximum rate of change of one variable with respect to another variable
terms of an AP is 1/v and the vth term is 1/u show that the sum of uv terms is 1/2(uv+1)
what is calculus?
calculus is math that studies the change in math, such as the rate and distance,
Tamarcus
what are the topics in calculus
Augustine
what is limit of a function?
what is x and how x=9.1 take?
what is f(x)
the function at x
Marc
also known as the y value so I could say y=2x or f(x)= 2x same thing just using functional notation your next question is what is dependent and independent variables. I am Dyslexic but know math and which is which confuses me. but one can vary the x value while y depends on which x you use. also
Marc
up domain and range
Marc
enjoy your work and good luck
Marc
I actually wanted to ask another questions on sets if u dont mind please?
Inembo
I have so many questions on set and I really love dis app I never believed u would reply
Inembo
Hmm go ahead and ask you got me curious too much conversation here
am sorry for disturbing I really want to know math that's why *I want to know the meaning of those symbols in sets* e.g n,U,A', etc pls I want to know it and how to solve its problems
Inembo
and how can i solve a question like dis *in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
next questions what do dy mean by (A' n B^c)^c'
Inembo
The sets help you to define the function. The function is like a magic box where you put inside stuff(numbers or sets) and you get out the stuff but in different shapes (forms).
I dont understand what you wanna say by (A' n B^c)^c'
(A' n B (rise to the power of c)) all rise to the power of c
Inembo
Aaaahh
Ok so the set is formed by vectors and not numbers
A vector of length n
But you can make a set out of matrixes as well
I I don't even understand sets I wat to know d meaning of all d symbolsnon sets
Inembo
High-school?
yes
Inembo
am having big problem understanding sets more than other math topics
Inembo
So f:R->R means that the function takes real numbers and provides real numer. For ex. If f(x) =2x this means if you give to your function a real number like 2,it gives you also a real number 2times2=4
pls answer this question *in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
If you have f:R^n->R^n you give to your function a vector of length n like (a1,a2,...an) where all a1,.. an are reals and gives you also a vector of length n... I don't know if i answering your question. Otherwise on YouTube you havr many videos where they explain it in a simple way
I would say 24
Offer both
Sorry 20
Actually you have 40 - 4 =36 who offer maths or physics or both.
I know its 20 but how to prove it
Inembo
You have 32+24=56who offer courses
56-36=20 who give both courses... I would say that
solution: In a question involving sets and Venn diagram, the sum of the members of set A + set B - the joint members of both set A and B + the members that are not in sets A or B = the total members of the set. In symbolic form n(A U B) = n(A) + n (B) - n (A and B) + n (A U B)'.
Mckenzie
In the case of sets A and B use the letters m and p to represent the sets and we have: n (M U P) = 40; n (M) = 24; n (P) = 32; n (M and P) = unknown; n (M U P)' = 4
Mckenzie
Now substitute the numerical values for the symbolic representation 40 = 24 + 32 - n(M and P) + 4 Now solve for the unknown using algebra: 40 = 24 + 32+ 4 - n(M and P) 40 = 60 - n(M and P) Add n(M and P), as well, subtract 40 from both sides of the equation to find the answer.
Mckenzie
40 - 40 + n(M and P) = 60 - 40 - n(M and P) + n(M and P) Solution: n(M and P) = 20
Mckenzie
thanks
Inembo
Simpler form: Add the sums of set M, set P and the complement of the union of sets M and P then subtract the number of students from the total.
Mckenzie
n(M and P) = (32 + 24 + 4) - 40 = 60 - 40 = 20
Mckenzie