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@Apu
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.As we apporach 2 and 3, the neighbourhood will consist of both rational and irrational
But the function is piecewise?
@Opt
how do i apporach this
I just woke up man, let me drink my coffee at least
Mb
i saw you were online and thus pinged you
sorry bro
Is it a
For approaching 2+, 2- and 3+, 3- consider f(x) = x^2+6
why
I think that the limit will only exist when x^2 + 6 = 5x
From the graph or by calculating the roots we can see that the fxn is continuous at x = 2 and x =3 is this right? Yup, the ans is (a) option
From the graph or by calculating the roots we can see that the fxn is continuous at x = 2 and x =3 is this right? Yup, the ans is (a) option
The graph of this function will contain loads of points (discontinuous). So graph analysis is difficult to comprehend. In competitive exams at secondary school level, what we do is we tryto put the value at which we wanna check continuity in both parts of function and see if we get the same value (if yes, then the function is continuous). If you notice, x=2 and x=3 will give the same value in both rational and irrational definitions, thus we have a limiting value at both points, since the definitions are polynomials, we will have continuity at those points. (Long answer will be a rigourous proof and mostly not required in exams in like JEE)
Alright sir. Thanks
+solved @Satya S
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