CSIR NET MATHEMATICS | Real Analysis | Uniform Convergent - MN Test, Solutions June 2019

Hello friends, in this video we are going to discuss the problem of CSI in June 2019. Example number 22 from booklet C marks 3. Let gnx is equal to nx divided by 1 plus n square x square. x belongs to close open integral 0 to infinity. Which of the following is true for n approach to infinity? First option gn converges to 0 pointwise but not uniform. Second option, Gn converges to 0 uniform. Third option, Gn converges to x for all x belongs to close interval 0 to infinity. And the fourth option, Gnx converges to x divided by 1 plus x square for all x belongs to 0 to infinity. So, here we will recall two important concepts. Convergence of sequence of function pointwise and uniform. So, first we will recall the definition of pointwise convergence. let D be a subset of R and let fn be a sequence of functions defined on D. We say that the convergence pointwise on D if limit n tends to infinity fnx exists for each point x in D. So this limit limit n tends to infinity fnx exists then we can say that fn is a pointwise convergent. Note that limit n tends to infinity fnx is equal to fx. This is pointwise limit of sequence fn. Now uniform convergent limit n tends to infinity fn equal to f uniformly convergent means limit n tends to infinity is supremum of modulus fn x minus fx where fx is a pointwise limit of fn equal to 0. To check the uniform convergent we will use mn test for uniform convergent. First step find limit n tends to infinity fnx is equal to fx that means point wise limit then the modulus fnx minus fx say hx calculate maximum of hx to find the maximum of hx just find extremum point by solving h dash x equal to 0. Suppose x0 is a point of extremum then calculate h double dash x at x0 if it is less than then hx has maximum value if limit n tends to infinity at the point of maximum this hx has maximum value 0 as n approach to infinity then it is uniform convergent otherwise it is not uniformly convergent. So for our example g nx is equal to nx divided by 1 plus n square x square. So in the first step we are going to calculate the pointwise limit limit n tends to infinity g n x is equal to limit n tends to infinity n x divided by 1 plus n square x square divide numerator and denominator by n square then it is limit n tends to infinity x divided by n 1 upon n square plus x square as n tends to infinity 1 upon n tends to 0 so numerator is 0 this is 0 plus x square so it is always 0 so g x which is a pointwise limit has value 0 for all x and limit exists therefore it is a point wise convergent but the third option gn converges to x for all x belongs to close open interval 0 to infinity it is wrong as well as fourth option is also wrong which says that gnx converges to x divided by 1 plus x square for all x because it will give the non-zero value third option and fourth option is cancelled now we are going to check the uniform convergent so for uniform convergent we are going to calculate the maximum of hx where hx is equal to gnx minus gx here gx is equal to 0 so its value is gnx and gnx is equal to nx divided by 1 plus n square x square so differentiation of h h dash x n minus n cube x square divided by 1 plus n square x square whole square now for the point of extremum h dash will be 0 so evaluate with 0 then we have n minus n cube x square is equal to 0 so after the simplification we get the value x is equal to 1 by n so x is equal to 1 by n is a point of extremum now calculate h double dash x by differentiation h dash x one more time and we have h double dash x is equal to minus 6 n cube x plus 2 n power 5 x cube divided by 1 plus n square x square power 3 now evaluate h double dash x at 1 by n then we have minus n square by 2 which is less than 0 therefore this point of extremum is a point of maximum so x equal to 1 by n is a point of maximum evaluate hx at point of maximum x is equal to 1 by n then limit n tends to infinity h of 1 by n substitute x is equal to 1 by n so n into 1 by n then numerator is 1 and the denominator of 1 plus n square into 1 upon n power 2 so n square n square cancel we have 1 so 1 upon 1 plus 1 which is 1 by 2 not equal to 0 therefore this limit has non-zero value therefore it is not uniformly convergent so it is pointwise convergent but not uniform which is option number one so option two is cancel and option one is correct answer thanks for watching for more videos like share subscribe press the bell icon of NextGen channel