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Very helpful 😊

Book name plzzz

Good explanation sir

Very nice lecture👍

Thank you

Excelente, consulta como seria el procedimiento para el limite (x^(1/n)-a^(1/n))/(x-a). Saludos

Aur please koyi notes waghera provide kare

Please more videos ,,daily

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But,like you said, we need to look every subcover of A. Just finding some subcover which is not finite doesn't h-show that R is non-compact. Cuz, a space is compact if every open cover has a finite subcover. There might be some subcovers of A such that they are finite.

very helpful and easy to understand. u should buy a mic so that your voice is more clear

assalamoalikum sir g Left almost semi group ka notes ya book mil jayega

What a trick to find all possible topology!! Great and astonishing work keep it up sir

7:41 X is compact, so it is a compact neighbourhood of x, no need to consider any open cover

Thank you 😊

This is the definition of Hausdorff space. Isn’t it? T1 is different from Hausdorff but this definition is for T1

thanks

Fix your sound first

Mashallah..

Topology ka chapter 8 ka 8.5.1therom k dono parts plzz upload kr dy

Thank you from India

Closed subbase meaning plz send definition of this if ur watching i need definition today only

GOD BLESS MORE SIR

plss check de morgans law.

Jazak-llahu khairan kasiran

Thak you so much ❤

Very nice explanation Sir 🙏

Sir m=max(m1,m2) q lia 02:38

Great vid

Nice 🎉

Very helpful

Good effort sir

Assalamualaikum..sir aap product spaces nd functional space py videos bnaen

Thanks for uploading this video. Every LA-Semigroup and RA-Semigroup holds medial or bi-symmetry law but the converse may not be true. Hope that you will upload more videos.

Nice problem but it would be so much nicer if you could use a word processor to show all the steps. And then make it downloadable. Three blue one brown has really nice presentation and the software is open source

1st question you wrote 64/2 = 2???

You made it simple sir! Be blesed

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❤❤❤❤❤❤❤❤❤❤❤❤❤ thanks sir

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Thankyou😄

Nice way

How can it be negative when you are taking absolute value????

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Good ❤❤❤❤❤❤❤🎉🎉🎉❤❤❤

mashallah sir slay inshallah

Well

Sir agr 0 to 2 ki jgh limit me 0 to 1 diya ho to bhi yhi solution krskte hai na

Thank you sir🥰

This is a typical case in which one should not use the definition of the derivative directly.Assuming the product rule is known we set y(x) = x^(1/3) so that y(x)^3 = x. Now use the product rule twice to see that (y(x)^3 )'= 3 y(x) ^2 * y '(x) . But (y(x) ^3) ' = x ' =1,and therfore y '(x) = 1/3 * 1/y(x)^2 = 1/3 x^(-2/3)