PreCalculus | Limits at Infinity

PreCalculus class on limits of a function at Infinity

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[cgreat]

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 ❤❤

[omeketoochukwulivinus]

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

[TambaEnsah]

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

[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 <1?

[imangbadamosi]

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

[Kennyluvdata]

Which no difference in solving for positive infinity and negative infinity

[adebayooluwaferanmiabati]

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?

[ngongfranklinesa-am]

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

[HajiKemal]

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

[HajiKemal]

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

[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?

[primuscliff]

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

[HajiKemal]

Why put 0 for each numerator in case evaluate each limit