Great sir ...... I enjoy when you are solving questions within 15 sec. in almost every lecture. Thanks sir ❤❤
@shikhasarkar4552
3 ай бұрын
13:52 tan(3) to tan(8) is infinite for atleast one values as 3+(π/2)
@priyanjalitripathi4408
9 ай бұрын
great sir......Thankyou so much for giving all concept of bounded variation in one video.
@amankumarbehera6232
2 күн бұрын
It is very helpful for us sir... keep it up,we are with you ❤😊🙏
@sarojsi890
9 ай бұрын
Good explanation with good collection 👍👍
@professorragavan9783
2 ай бұрын
Thankyou very much sir. I did not know about a single concept of bounded variation. After watching your video, it is very useful to me. 🤩😍
@DrHarishGarg
2 ай бұрын
You are most welcome
@ashokisaini8427
9 ай бұрын
Superb 👍👍👍
@VinayTiwari-hp1pp
9 ай бұрын
Nice explanation sir
@KAREN-ye5pf
3 ай бұрын
27.22 Problem (b) x=1 is not in the domain itself how can we find tan (pi/2)=infinity and we say it's not BV sir ?
@ChandanKumar-dn9lz
9 ай бұрын
Excellent explanation... Outstanding....in just one hr lecture each and every thing got Cristal clear ....thanks a lot sir jee.
@user-lo2qx8ow9e
9 ай бұрын
Your explanations and the way you clear the concept is just wow...❤
@MANTUKUMAR-nn8kk
3 ай бұрын
Gajab sir ❤❤❤
@RS10811
9 ай бұрын
❤❤❤ eske alawa kya hi likh den...kyuki sir se accha samjhane wala Aaj Tak dusra koi Mila nhi. Thank you sir. 🙏
@Prayosi
8 ай бұрын
These lectures are very good. Thank you so much sir.
@ManishMeena-eo4qh
8 ай бұрын
Bht bht dhanyvad guru ji 😊😊😊😊
@ranusheoran3303
9 ай бұрын
Super explanation 👍
@yashrajtripathi4832
Ай бұрын
Apke lecture se pura topic revise ho jata + pyq to ho hi jate h
@jkmathematics..jitendrakoh8557
3 ай бұрын
Thanku so much sir love from ❤❤ uttarakhand
@madhavigandhi3181
6 күн бұрын
NBHM 2014 For last option sir, we can say f is bounded by f(0) and below by f(1) therefore it is bdd.
@nehajangra6362
4 ай бұрын
Mistakes in last bouned variation implies monotonicity Monotonic + bdd is b.v. but ek nbhm k question main closed interval per monotonic diya hua h how it implies that function is of b.v.
@positivethoughts6419
9 ай бұрын
Thank you so much for this important video 🙏🏼...
@vivekgoswami474
9 ай бұрын
Thanks sir , next complex pleasr
@sadzone1434
9 ай бұрын
Nice
@rohitgoswami7812
9 ай бұрын
Thank you sir.❤
@selfsearching1996
9 ай бұрын
Nice 👍👍
@madhavigandhi3181
5 ай бұрын
Thank You very much sir, for guiding us selflessly. Your videos are marvellous ! Very useful, clear and doubt solving.
@kunjkothiya7990
3 ай бұрын
At time 34:10. In this question A is correct option.
@somduttakayal5067
9 ай бұрын
Thank you Sir
@yashrajtripathi4832
Ай бұрын
Thank you sir 😊
@jaikmthomas5035
9 ай бұрын
@50:00 Sir if function is not continuous then need not be f is not reimann integrable
@jyotibharti6505
9 ай бұрын
Thank you so much sir 🙏🙏🙏🙏
@mathematicslovers2888
2 ай бұрын
Thank you sir for giving us very good Content ❤❤
@DrHarishGarg
2 ай бұрын
My pleasure... Kindly like, subscribe and share the video with others students too...
@mathematicslovers2888
2 ай бұрын
Sure sir ☺️ @@DrHarishGarg
@vishal-ez8tz
6 ай бұрын
at 49:41 function is not continuous only at 1 . Therefore discontinuity at only finite number of points this implies f is riemann integerable .Am i right? plz verify sir
@user-uv3ef6ps2t
2 ай бұрын
Very nice sir Kindly give on continous and differentiability uniform continuity topic sir before june 2024 csir pleaseeee
@positivethoughts6419
9 ай бұрын
Sir... Please this type of one by one topic video with short explanation of complex analysis subject...🙏🏼
@steel679
9 ай бұрын
Thankyou so much sir,best youtube teacher i ever seen, ❤❤❤❤❤❤keep it sir, short time :=>max knowledge = Dr. HARISH GARG❤
@parijatnath4765
9 ай бұрын
Sir please upload an pyqs video on lebsigue integration
@ketemawdemeke7706
9 ай бұрын
You email is not active i want to ask one question
@easymathematics1341
9 ай бұрын
❤
@sanjitha_dhivin
4 ай бұрын
Sir if we get a function that asymptotically decreases at 1. how the function will become bounded on [0,1]?
@vikashgupta-xw5np
5 ай бұрын
❤❤❤
@sanjitha_dhivin
4 ай бұрын
Sir how sinx becomes monotonous on [0,2phi]?
@keyproblemsinhighermathema1778
5 ай бұрын
If f is increasing on [a,b] then f is BV?
@techmath8922
9 ай бұрын
Really very happy ... thank you for your great effort
@memebaaz4005
8 ай бұрын
48.59 min I’m confused how pointwish congt because it ls jo continuous
@shrawankumarpatel2148
9 ай бұрын
🙏🌷🌷🌷🌷🙏🌷🌷🙏
@hajeeranasreenqueenofscience
9 ай бұрын
I have a doubt sir For the pointwise or piecewise continuous,is it enough to check limit exits or it must be both , limit exists and are same ?
@khushiracca4986
4 ай бұрын
just limit exists is enough....not necessary it should be same
@nishayadav9717
9 ай бұрын
Why x is not of BV over (-♾️,♾️)....x is lipschitz over (- ♾️, ♾️) so x is of BV... please correct it sir
@DrHarishGarg
9 ай бұрын
Here domain is unbounded , w
@nishayadav9717
9 ай бұрын
@@DrHarishGarg sir x is of BV over (- ♾️, ♾️)
@nishayadav9717
9 ай бұрын
Because derivative bounded throughout the domain
@ankita-11168
6 ай бұрын
@@nishayadav9717 as you say x is bv over (-infinity, infinity) then it implies that x is bounded over (-infinity, infinity) but x is not bdd over this domain
@nehajangra6362
4 ай бұрын
Plzzzz sir phle kud confirm krke bad main hume bataye u r little bit confused
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