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

  Рет қаралды 4,451

Dr. Mathaholic

Dr. Mathaholic

3 жыл бұрын

Пікірлер: 18
@saumitradeo5336
@saumitradeo5336 3 жыл бұрын
Sir, In the first example, for DCT, if theta=5pi/6, then the inequality for the last two wouldn't hold.
@DrMathaholic
@DrMathaholic 3 жыл бұрын
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
@nighatlone77 3 жыл бұрын
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
@DrMathaholic 3 жыл бұрын
Closed set
@nighatlone77
@nighatlone77 3 жыл бұрын
@@DrMathaholic thank u sir....
@helloworld-hv9oy
@helloworld-hv9oy 3 жыл бұрын
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
@DrMathaholic 3 жыл бұрын
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
@helloworld-hv9oy 3 жыл бұрын
@@DrMathaholic okay sir thank you !
@legendgaming8020
@legendgaming8020 2 жыл бұрын
Sir how we solve this sum>>>>. 1/(1+x)x^1/2 converge or devearge i
@legendgaming8020
@legendgaming8020 2 жыл бұрын
Limit x =0 to infinity
@DrMathaholic
@DrMathaholic 2 жыл бұрын
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.
@SumitGaming-
@SumitGaming- 2 жыл бұрын
Sir in 3rd part 9:29 ,it converges
@DrMathaholic
@DrMathaholic 2 жыл бұрын
How? 🤔🤔🤔 1/x^3 is diverging right?
@SumitGaming-
@SumitGaming- 2 жыл бұрын
@@DrMathaholic sorry .you are right
@DrMathaholic
@DrMathaholic 2 жыл бұрын
@@SumitGaming- 🙌🙌🙂
@krishnadehrawala442
@krishnadehrawala442 2 жыл бұрын
Sir we can apply any test from limit and direct ?
@DrMathaholic
@DrMathaholic 2 жыл бұрын
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
@krishnadehrawala442 2 жыл бұрын
@@DrMathaholic ok
Lecture 32: Improper Integrals of Type II. Concept and Examples.
16:13
ВОДА В СОЛО
00:20
⚡️КАН АНДРЕЙ⚡️
Рет қаралды 32 МЛН
Smart Sigma Kid #funny #sigma #comedy
00:40
CRAZY GREAPA
Рет қаралды 23 МЛН
What it feels like cleaning up after a toddler.
00:40
Daniel LaBelle
Рет қаралды 86 МЛН
Divergence or Convergence of Improper Integrals
27:31
Prime Newtons
Рет қаралды 6 М.
Limit comparison and direct comparison test (improper Integrals)
30:44
Hoa Kim (FVHS)
Рет қаралды 2,3 М.
Steven Strogatz: In and out of love with math | 3b1b podcast #3
1:54:08
Grant Sanderson
Рет қаралды 209 М.
The Most Beautiful Equation in Math
3:50
Carnegie Mellon University
Рет қаралды 13 МЛН
Comparison test for convergence and divergence of improper integrals
16:56
bprp calculus basics
Рет қаралды 71 М.
Riemann sum examples (calculus 1 limits vs integrals)
22:50
bprp calculus basics
Рет қаралды 21 М.
Series | Lecture 18 | Condition for Convergence of p - series
12:45
Limit Comparison Test Integrals - Example
6:34
JCCCmath
Рет қаралды 9 М.
ВОДА В СОЛО
00:20
⚡️КАН АНДРЕЙ⚡️
Рет қаралды 32 МЛН