indian students in 11 and 12 practically getting their math knowledge redefined here✊
@-2025S
Жыл бұрын
Did you crack JEE or NEET?
@abhinavbadoni7032
Жыл бұрын
I wish i had this opportunity when i was in class 11 12
@jadon7878
6 жыл бұрын
cheers got a test tomorrow and you may have saved my life :)
@gsridevi4495
4 жыл бұрын
I want to be like you in speaking and teaching and your approachings sir.
@madannavali3959
3 жыл бұрын
8:35 yeah sounds soooo good
@PoppySuzumi1223
3 жыл бұрын
So can I say that if a function / a graph has one or more asymptotes, then this graph must not have a continuity ?
@carultch
Жыл бұрын
Depends on what kind of asymptotes you are talking about. A vertical asymptote at a finite x-value, approached on both sides, yes. But a horizontal asymptote approached on only one side, not at all.
@studyplusmathematics562
4 жыл бұрын
Watching your video 2nd time u r awsome
@johannasvensson995
5 жыл бұрын
But x=0 is not even in the domain of the function y=1/x, so in fact, y=1/x is a continuous function.. right?
@hrishikeshsarma1796
5 жыл бұрын
Yup, it's domain is all Real number except zero, but here we try'in to grasp the formal definition of continuity i.e. LHL = RHL = value of func. at that point, which is in this case the limit does not exist, so we don't go further to check for f(a) ;)
@randygodi432
3 жыл бұрын
The limit wont exist becuase the limits from the positive and negative sides arent equal. Thus it is not continius
@billyjoethethird8436
5 жыл бұрын
Why does the continuity test have to include both the positive and the negative limits? Why not just plug in the value and see if it comes up as a real answer? Because if it is solvable, then the limit should be able to approach it from both sides right?
@hrishikeshsarma1796
5 жыл бұрын
U're right! For a func. that is solvable, we can just plug in the value into the func. But as u said the func is solvable!. Now, what do we do if it is unsolvable i.e. for some value it is undefined, then we can check by using the above mentioned left and right hand limits, or factorizing or rationalizing & stuff.. ;)
@ifg.alphawolf1675
4 жыл бұрын
@@hrishikeshsarma1796 Incorrect! The limits exists if the function value equals both the right and left side limits. So just because the function is defined this doesn't mean that the limit exists. Look up some examples on google (piecewise functions especially).
@carultch
Жыл бұрын
Because there are functions defined along an interval at all x-points along it, but that are not continuous. You can plug in a value at every point along the way, but you can get a wildly different value if you plug in its immediate neighbor.
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