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
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[@m.isworld2562]

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]

Thank you very much!

[@bprpcalculusbasics]

Could someone please help with the time stamps? It’s late and I haven’t gotten much sleep lately. Thank you.

[@angelvargas9042]

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

[@nonentity168]

We are blessed to have online educators like you. Thanks a lot

[@papkenhartunian186]

One of the best lectures that I have ever attended. Bravo!

[@ChrisP999]

Absolute wonderful sight, so glad to many examples to see different approaches to these. Keep it up!

[@sarmadmaqsood727]

Thank you for this! Now I have more practice using L'Hopitals!

[@dpmike32819]

Excellent selection of problems

[@muslimguy1068]

Thanks for the explanation man!

[@mikejackson19828]

Great video, Steve!

[@user-db6zu2gv2t]

Thank you for this fine piece if art in maths

[@margot947]

Thank you, so much for sharing

[@oyebi2553]

great video thanks !

[@ruchigaur5127]

Perfect video

[@eduardvalentin830]

18:52 sub u=1/x-1 and you get the definition of e. Nice!

[@isharauditha4257]

you are the saviour

[@jayprados1089]

This man is the son of the old guy from Harry Potter

[@gigathus3314]

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!

[@boudescotch]

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]

Me too, it actually slowly rises after dipping at around x=20.086

[@Shawon00s]

Thank you...

[@KgopotsoMoloto-vx2rq]

Thank u sir😁😍

[@sohanurrahman1515]

Thanks sir

[@turkix4809]

How theres only 57K viewers You are a world class teachee

[@acopiae]

and just 700 likes. must be too niche

[@gmoney_swag1274]

his main channel has millions of subscribers (blackpenredpen)

[@danglersolutions2]

1/x - 1/sinx as x approaches 0

[@sri78hari]

Slowly you became Confucius by teaching math😂

[@chananleibler6901]

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]

The ideia is that the 1+ can be closer of 1 in a way "more stronger" than the infinity

[@nelsonberm3910]

you are a mega g

[@abdullahkamel2428]

i fucking love this guy

[@ires3731]

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]

presumably to keep the number positive, otherwise you have the log of a negative number.

[@gileadedetogni9054]

That's because you're approaching positive infinity, so you get "0+"

[@JoyJohn-ux3cm]

I have assignment, can u help me?

[@boonga585]

5:41

[@pogisomakaba4829]

Isn't number 7 supposed be e^-1

[@hydropage2855]

Professor Chow, the worksheet link was taken down. Just wanted to inform you

[@bprpcalculusbasics]

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.

[@woshibaba428]

I love you

[@omarhmidani6063]

2:00 just seen the function and found the solution

[@danglersolutions2]

please can you help me with this maths

[@jimmycatyen]

Question 12 We can solve it by looking 3-0/2=3/2

[@aditidump]

can u pls xplain further?

[@boonga585]

26:39

[@Cycy98MinecraftFR-xh9lr]

I think it's also spell L'Hôpital's rule, I think it because hopital seem French.

[@walidslimani]

there is a lot you can solve them without l'hopitals rule

[@petereziagor4604]

Integration rules please

[@pardeepgarg2640]

Lopital's Rule 👁️👄👁️ YaY

[@jannawang9279]

why do you have a pokemon ball?

[@MuhammadBilal-pi3lo]

Because his brain power of Maths lies there

[@kittynpaws6580]

It's his mic in disguise lol

[@user-jz8fp3su7c]

just have fun with math for around 45 minutes

[@Azuxs]

Number 13 is 0, not infinity.

[@bprpcalculusbasics]

It's infinity. www.wolframalpha.com/input?i=limit+of+e%5Ex*sin%281%2Fx%29+as+x+goes+to+inf

[@Azuxs]

@@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.

[@nightmarefredbear7739]

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

[@hipolitonovido8651]

Too fast for me lol

[@solitary6002]

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]

The classic engineer approach 🤣 but try it for 1/sin(x)^2 - 1/x^2

[@ZipplyZane]

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.