Lecture 34: Limit Comparison Test for Improper Integrals of Type I and Type II.(see pinned comment)

[@saumitradeo5336]

Sir, In the first example, for DCT, if theta=5pi/6, then the inequality for the last two wouldn't hold.

[@DrMathaholic]

Yes, you are correct dude.. i was bit casual 😔 its only correct till 0 to π/2 and from π/2 to π inequality reverses... So the best way is substitute π-theta= x.

[@nighatlone77]

The set [a, b] belogs to [c, d] is A) closed set B) open set. (C) nether closed nor an open set (D) neither an interval nor an open set

[@DrMathaholic]

Closed set

[@nighatlone77]

@@DrMathaholic thank u sir....

[@helloworld-hv9oy]

Sir for the last questions is that test required ,beacuse we can directly evaluate that integral ? One more doubt ,it seems that for the limit test the convergence and divergence of f(x) depends upon the function g(x) we choose ...so can it be possible that there are two such functions and different answers come ?

[@DrMathaholic]

If you can evaluate directly then no problem at all. One can choose different different g(x) no problem in that, but the good thing is nature of f(x), whether it convergent or divergent, does not changes.

[@helloworld-hv9oy]

@@DrMathaholic okay sir thank you !

[@legendgaming8020]

Sir how we solve this sum>>>>. 1/(1+x)x^1/2 converge or devearge i

[@legendgaming8020]

Limit x =0 to infinity

[@DrMathaholic]

rewriting as: root(x)/(x(x+1)) = root(x)(1/x - 1/(x+1)) = root(x)/x - root(x)/(x+1) . for first term, split the integration 0 to 1 and 1 to infinity. for 1 to infinity 1/root(x) diverges. so i think answer is divergent.

[@krishnadehrawala442]

Sir we can apply any test from limit and direct ?

[@DrMathaholic]

Yes , just check the conditions..if the functions satisfies the conditions in that theorem then you can apply the respective test. If one test fails then try for second one..

[@krishnadehrawala442]

@@DrMathaholic ok

[@SumitGaming-]

Sir in 3rd part 9:29 ,it converges

[@DrMathaholic]

How? 🤔🤔🤔 1/x^3 is diverging right?

[@SumitGaming-]

@@DrMathaholic sorry .you are right

[@DrMathaholic]

@@SumitGaming- 🙌🙌🙂