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.
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.
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 ?
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.
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..
Sir, In the first example, for DCT, if theta=5pi/6, then the inequality for the last two wouldn't hold.
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.
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
Closed set
@@DrMathaholic thank u sir....
Sir in 3rd part 9:29 ,it converges
How? 🤔🤔🤔
1/x^3 is diverging right?
@@DrMathaholic sorry .you are right
@@SumitGaming- 🙌🙌🙂
Sir how we solve this sum>>>>. 1/(1+x)x^1/2 converge or devearge i
Limit x =0 to infinity
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.
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 ?
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.
@@DrMathaholic okay sir thank you !
Sir we can apply any test from limit and direct ?
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..
@@DrMathaholic ok