L'Hopital's Rule ultimate study guide
Ultimate calculus tutorial on how to use L'Hopital's Rule (also spelled as L'Hospital's Rule) to evaluate limits with indeterminate forms? In this calculus tutorial, we will do 14 examples of using L'Hoptial's Rule to evaluate limits with indeterminate forms. Remember we can only use L'Hospital's Rule only if we have 0/0 or infinity/infinity. So if we encounter other indeterminate forms (i.e. 1^inf, 0^0, infinity - infinity, 0*infinity, inf^0), then we must somehow change the limit indeterminate forms first before using L'Hospital.
Try the problems first 👉 936933f9-1455-44ce-b414-4d0b3...
Big thanks to M.I's World for the following timestamps:
0:00 start of the video
0:07 x-sinx/x-tanx as x goes to 0
3:16 (arctanx)^2/x as x goes to 0
4:32 ln(x^3-8)/ln(x^2-3x+2) as x goes to 2+
9:45 x/(lnx)^3 as x goes to infinity
13:13 x^2+4x+3/5x^2-x-4 as x goes to infinity
15:48 ln(e^x+1)/(4x+1) as x goes to infinity
18:52 x^(1/(x-1)) as x goes to 1+
24:05 x^(1/(1+lnx)) as x goes to 0+
27:08 (x/(x-1)-1/lnx) as x goes to 1+
31:55 (1/sinx - 1/x) as x goes to 0+
**Q10. behind the scene: • Video
34:36 ln(x^2-1)-ln(x^3-1) as x goes to infinity
36:42 sqrt(x^2+3x+1)-x as x goes to infinity
39:23 e^x times sin(1/x) as x goes to infinity
43:47 xarctan(x) as x goes to infinity
---------------------------------------------------------
This channel is dedicated to students who are taking precalculus, AP calculus, GCSE, A-Level, year 12 maths, college calculus, or high school calculus. Feel free to leave calculus questions in the comment section and subscribe for future videos.
Subscribe by clicking 👉 bit.ly/just_calc
---------------------------------------------------------
If you find my channel helpful and would like to support, then please check out
My Patreon: 👉 / blackpenredpen
My merch store: 👉 bit.ly/bprp_merch
Пікірлер: 64
0:00 start of the video 0:07 x-sinx/x-tanx as x goes to 0 3:16 (arctanx)^2/x as x goes to 0 4:32 ln(x^3-8)/ln(x^2-3x+2) as x goes to 2+ 9:45 x/(lnx)^3 as x goes to infinity 13:13 x^2+4x+3/5x^2-x-4 as x goes to infinity 15:48 ln(e^x+1)/(4x+1) as x goes to infinity 18:52 x^(1/(x-1)) as x goes to 1+ 24:05 x^(1/(1+lnx)) as x goes to 0+ 27:08 (x/(x-1)-1/lnx) as x goes to 1+ 31:55 (1/sinx - 1/x) as x goes to 0+ 34:36 ln(x^2-1)-ln(x^3-1) as x goes to infinity 36:42 sqrt(x^2+3x+1)-x as x goes to infinity 39:23 e^x times sin(1/x) as x goes to infinity 43:47 xarctan(x) as x goes to infinity im not sure if these are all correct timestamps
@bprpcalculusbasics
2 жыл бұрын
Thank you very much!
Could someone please help with the time stamps? It’s late and I haven’t gotten much sleep lately. Thank you.
Its crazy watching this guy and the organic chemistry guy. Currently studying for my engineering license and need to watch these videos to remember basic cal properties. I remember watching you and the organic chemistry guy 7 years ago at the junior college. Its such a trip to still watch your videos many years later. I remember struggling so bad on this stuff and after just completing reinforced concrete design. I realize that it wasn't so bad. I guess what I'm trying to say is that as the years go on, your going to realize that this stuff wasn't so bad and you'll just honestly laugh at yourself lol. This is the reason why many people don't go to college because with each new semester, comes new emotions, new struggles and most importantly a new version of you. Each semester we all have ever taken continues to add more value to your life and increase your critical thinking skills. You may or may not use calculus for your career but the critical thinking skills will reveal itself when you need it the most. Don't give up on your educational journey. Your future self is waiting for you on the other side. - The engineer turned PreMed
We are blessed to have online educators like you. Thanks a lot
Absolute wonderful sight, so glad to many examples to see different approaches to these. Keep it up!
One of the best lectures that I have ever attended. Bravo!
Thank you for this! Now I have more practice using L'Hopitals!
Excellent selection of problems
Thanks for the explanation man!
Great video, Steve!
Thank you for this fine piece if art in maths
Thank you, so much for sharing
Perfect video
18:52 sub u=1/x-1 and you get the definition of e. Nice!
you are the saviour
Thanks sir
Thank you...
Thank u sir😁😍
This man is the son of the old guy from Harry Potter
Slowly you became Confucius by teaching math😂
How theres only 57K viewers You are a world class teachee
@acopiae
Жыл бұрын
and just 700 likes. must be too niche
@gmoney_swag1274
Жыл бұрын
his main channel has millions of subscribers (blackpenredpen)
1/x - 1/sinx as x approaches 0
After the x/(lnx)³ I goes look the graph of the function and this is pretty surprising that the limit as x goes to infinity is infinity
@memebaltan
4 ай бұрын
Me too, it actually slowly rises after dipping at around x=20.086
i fucking love this guy
you are a mega g
In number 7 isn’t 1+^inf always infinity, since if you think about it 1+ ≈1.0000…1 and if you multiply that by itself infinite times it will approach infinity? And similary wouldn’t 1-^inf approach 0?
@gileadedetogni9054
Ай бұрын
The ideia is that the 1+ can be closer of 1 in a way "more stronger" than the infinity
Wonderful video man, thank you so much for the help! I do have a question though, in regards to the problem at 27:08 - Why couldn't we simply do L'Hopital's Rule to both ends of the difference instead of simplifying and making an equal denominator? If we do L'Hopital's Rule to both ends of the difference, we get that the Limit is equal to "0" instead of 1/2. I'm wondering why this doesn't work because from what I remember, if you are taking the limit of an expression that includes addition and/or subtraction, you can split the expression into the sum of limits like this: (lim x->1^+ (x/(x-1))) - (lim x->1^+ (1/lnx)). Why can't L'Hopitals Rule be applied like this? Thank you so much for the video you are saving so many students grades!
@npsiddhant3474
6 күн бұрын
If you apply the limit it becomes 1^(+)/1^(+) - 1 =a lets say and 1/ln(0^+)=b So a-b The a term becomes ∞ and the B term 1/(-∞) i.e 0 and then a-b becomes ∞-0
Question 12 We can solve it by looking 3-0/2=3/2
@aditidump
Жыл бұрын
can u pls xplain further?
5:41
Professor Chow, the worksheet link was taken down. Just wanted to inform you
@bprpcalculusbasics
2 жыл бұрын
Thanks for letting me know. The file is okay but I think many people have been finding bit.ly links not working for them. So here's the original link (long link, lol) 936933f9-1455-44ce-b414-4d0b35a6c090.filesusr.com/ugd/287ba5_fc19d8f3e1a94c4295298047578e2197.pdf and I also changed it in the description. Thanks.
I love you
for q11, when u bring the limit inside, why is it approaching 0 from the right? shouldn’t it just be 0 since it’s the limit from just infinity?
@acopiae
Жыл бұрын
presumably to keep the number positive, otherwise you have the log of a negative number.
@gileadedetogni9054
Ай бұрын
That's because you're approaching positive infinity, so you get "0+"
2:00 just seen the function and found the solution
Lopital's Rule 👁️👄👁️ YaY
26:39
there is a lot you can solve them without l'hopitals rule
I have assignment, can u help me?
Isn't number 7 supposed be e^-1
please can you help me with this maths
I think it's also spell L'Hôpital's rule, I think it because hopital seem French.
Integration rules please
just have fun with math for around 45 minutes
why do you have a pokemon ball?
@MuhammadBilal-pi3lo
7 ай бұрын
Because his brain power of Maths lies there
@kittynpaws6580
4 ай бұрын
It's his mic in disguise lol
Number 13 is 0, not infinity.
@bprpcalculusbasics
2 ай бұрын
It's infinity. www.wolframalpha.com/input?i=limit+of+e%5Ex*sin%281%2Fx%29+as+x+goes+to+inf
@Azuxs
2 ай бұрын
@@bprpcalculusbasics Could you explain further? How come after putting sin(1/x)/e^-x and you plug in infinity = sin(0)/infinity = 0/infinity = 0.
0:06 Problem 1 3:15 Problem 2 4:33 Problem 3 9:45 Problem 4 13:13 Problem 5 15:47 Problem 6 18:52 Problem 7 23:58 Problem 8 27:08 Problem 9 31:56 Problem 10 34:35 Problem 11 36:42 Problem 12 39:25 Problem 13 43:47 Problem 14
Why work so much for the question 10? x tends to 0+....sin x = x for a very very small angle... Thus 1/sin x = 1/x Thus lim x tends to 0+ 1/sin x - 1/x = 0 That's what I though on seeing the question 🤣
@isaquepim4555
2 жыл бұрын
The classic engineer approach 🤣 but try it for 1/sin(x)^2 - 1/x^2
@ZipplyZane
2 жыл бұрын
To be more explicit: this works, but only with linear terms. It doesn't work the second you raise sin(x) to any power. The reason it works is because of the Taylor expansion of sine. If you have an exponent of 1, all the other terms are insignificant. Michael Penn (the math teacher) has a video on this. It's called "When This Approximation Goes Wrong," and shows sin x ≈ x in the thumbnail.
Too fast for me lol