PreCalculus | Limits at Infinity

[@omeketoochukwulivinus4119]

Sir, i don't know but you're simply the best, how comes i haven't discovered you yet all this while? All of my major problems under calculus are just been handled here. Thank you for this great tutorial, God bless you, you're such a genius

[@jonahemmanuelofficial]

Awesome, thank you. Where are you watching from?

[@cgreat8983]

SIR THANK YOU SO MUCH I GOT AN A IN MATHS and you helped me do that. God bless you sir for uploading these videos SIMPLY THE BEST ❤❤

[@jonahemmanuelofficial]

Awesome. Happy to hear. Congratulations to you

[@TambaEnsah]

You made me become good in maths congratulations 🎉 sir.I love watching your videos ❤❤❤❤

[@jonahemmanuelofficial]

Awesome. I love giving you here. Thank you

[@mosesuwala5199]

Well done, Great job

[@jonahemmanuelofficial]

Thank you so much 👍

[@KatelaArarsa]

Thanks sir iam From Ethiopia

[@jonahemmanuelofficial]

Awesome. Welldone 👍

[@danielofuru2401]

Nice video very well understood 😊

[@user.chukwuladi00]

Good one Sir Jonah ❤

[@jonahemmanuelofficial]

Sure. Welcomed..!!!

[@thecreativehub88]

Thank you very much sir 😊

[@saraasare699]

Great job

[@jonahemmanuelofficial]

Well appreciated, where are you watching from?

[@mizlolia]

Nice one Investor J!!! Thank you. So, whether +ve or -ve infinity, the method is the same? And as far as infinity is concerned, our answer must be zero right? =0 but

[@jonahemmanuelofficial]

Nah, must not always be zero. Check out other examples too

[@ideraolubello4046]

Whether the infinity is -ve or +ve We use the same method right Ur classes are so amazing 🤩

[@jonahemmanuelofficial]

@ideraolubello4046 yes yes, thank you so much 👍

[@adebayooluwaferanmiabati7542]

Thank you very much for these lessons. For the first example, you left the answer as 0 and according to you, you said it’s undefined. Do we stil leave the answer as 0?

[@jonahemmanuelofficial]

Welcomed. Undefined is when denominator is zero. When numerator is zero, the Fraction becomes zero

[@imangbadamosi]

Does that mean the method of solving -ve infinity Is the same method for solving +ve infinity?

[@jonahemmanuelofficial]

Yes, it is.

[@TDSONLINEMATHS]

Great

[@jonahemmanuelofficial]

Sure.

[@N_Vamseedharan]

@jonahemmanuelofficial Great session sir. Could you clarify this to me? In this infinity method substituting x=∞ would be more logical than substituting x = 0. In example 2, after algebraic simplification, (1) Lim x-> 0 = {6-4/0³ +3/0⁵}/{11+2/0-1/0⁴} = which would again end up as infinity but if, (2) Lim x-> ∞ = {6-4/∞³ +3/∞⁵}/{11+2/∞-1/∞⁴} = {6-0+0}/{11+0-0} = 6/11 [as 1/0 = ∞] I feel the second (2) one is more logical and more correct. AM I RIGHT IN THIS ASPECT?

[@jonahemmanuelofficial]

Ok then

[@N_Vamseedharan]

@@jonahemmanuelofficial Thank you

[@ikyrie]

Legend

[@jonahemmanuelofficial]

Thank you ❣️

[@ElvisBoateng-q1u]

SHORT BUT WORTH LISTENING 👂 👂

[@jonahemmanuelofficial]

Yup

[@HajiKemal]

Why put 0 for each numerator in case evaluate each limit

[@Kennyluvdata]

Which no difference in solving for positive infinity and negative infinity

[@jonahemmanuelofficial]

Yesss.,none, basically.

[@franciscaoseinyarko1321]

pleasebwhat about the negative attached to the infinity

[@jonahemmanuelofficial]

I talked about it there. I said positive or negative infinity.... same thing

[@HajiKemal]

Sir 3÷0 = 0 is in correct in case you should have to take lim3/x^2 = 0

[@jonahemmanuelofficial]

Punch that on your calculator and see if it's correct. 0/3 = 0 3/0 = undefined (calculator will show an error)

[@HajiKemal]

Jonah at the end you put 0 why is 4÷0= 0

[@jonahemmanuelofficial]

4÷0 is not equal to zero 🙂

[@primuscliff7313]

Thanks sir . Please can you do a video on FUNCTIONS, involving Asymptote ( VA, HA, OA) Thanks

[@jonahemmanuelofficial]

Ok then.

[@ngongfranklinesa-am9414]

My worry is why dividing by 0 . Mathematicaly we don't divide by 0.

[@EmmanuelEssuman-lz5pu]

Learn it well

[@shekugbla5511]

👍❤️🇸🇱

[@KatelaArarsa]

Thanks sir iam From Ethiopia

[@jonahemmanuelofficial]

Welcome.