Show that lim n → ∞ ∫ a ∞ n 2 x e − n 2 x 2 1 + x 2 d x = 0 for a &gt; 0 but not for a = 0.My work: Let the value of the given limit be L . Put n 2 x 2 = t in the integrand. Then we have: (1) L = 1 2 ⋅ lim n → ∞ ∫ n 2 a 2 ∞ e − t 1 + t / n 2 d t . Now plugging in t = n 2 a 2 + z yields: L = 1 2 ⋅ lim n → ∞ ∫ 0 ∞ e − n 2 a 2 ⋅ e − z 1 + a 2 + z / n 2 d z .The last integrand is less than e − z , an integrable function in [ 0 , ∞ ) . Also as n → ∞ , the same integrand approaches 0. Hence by dominated convergence theorem, we obtain that L = 0.The way I have solved the problem, I am not noticing any effect of the sign of a on the value of L . However, if I consider a = 0 then from (1) using a similar approach I found that L = 1 2 . So sign of a does matter. So my main query is what is wrong with my approach? Please give some insights.Thanks in advance.EDIT. Another thing that came into my mind just now is the following:If we substitute n x = t , then: L = lim n → ∞ ∫ n a ∞ t e − t 2 1 + t 2 / n 2 d t = lim n → ∞ ∫ 0 ∞ χ ( n a , ∞ ) t e − t 2 1 + t 2 / n 2 d t , where χ A represents the characteristic function of the set A.Here it seems that if a &lt; 0 then it would affect the value of L and to be honest I'm not quite sure about the dominant integrable function anymore. Please shed some light on the same, particularly for the case a &lt; 0.

Question

Show that             lim              n        →        ∞                            ∫        a                  ∞                                                  n            2                    x                      e                          −                              n                2                                            x                2                                                              1          +                      x            2                                    d      x      =      0       for   a  &amp;gt;  0 but not for   a  =  0.My work: Let the value of the given limit be   L  . Put       n    2        x    2    =  t in the integrand. Then we have:                    (1)                    L        =                  1          2                ⋅                  lim                      n            →            ∞                                    ∫                                    n              2                                      a              2                                            ∞                                                e                          −              t                                            1            +            t                          /                                      n              2                                              d        t        .            Now plugging in   t  =      n    2        a    2    +  z yields:  L  =      1    2    ⋅      lim          n      →      ∞            ∫    0          ∞                          e                  −                      n            2                                a            2                              ⋅              e                  −          z                            1      +              a        2            +      z              /                    n        2                d  z  .The last integrand is less than       e          −      z        , an integrable function in   [  0  ,  ∞  )  . Also as   n  →  ∞  , the same integrand approaches 0. Hence by dominated convergence theorem, we obtain that   L  =  0.The way I have solved the problem, I am not noticing any effect of the sign of   a on the value of   L  . However, if I consider   a  =  0 then from (1) using a similar approach I found that   L  =      1    2    . So sign of   a does matter. So my main query is what is wrong with my approach? Please give some insights.Thanks in advance.EDIT. Another thing that came into my mind just now is the following:If we substitute   n  x  =  t  , then:                    L                            =                  lim                      n            →            ∞                                    ∫                      n            a                                ∞                                                t                          e                              −                                  t                  2                                                                          1            +                          t              2                                      /                                      n              2                                              d        t                                                        =                  lim                      n            →            ∞                                    ∫          0                      ∞                                    χ                      (            n            a            ,            ∞            )                                                t                          e                              −                                  t                  2                                                                          1            +                          t              2                                      /                                      n              2                                              d        t        ,            where       χ    A   represents the characteristic function of the set   A.Here it seems that if   a  &amp;lt;  0 then it would affect the value of   L and to be honest I&#039;m not quite sure about the dominant integrable function anymore. Please shed some light on the same, particularly for the case   a  &amp;lt;  0.

antantil0 · Accepted Answer

HintAnother way is the squeeze theorem:      ∫    a          ∞                          n        2            x              e                  −                      n            2                                x            2                                      1      +              x        2                d  x  ≤      ∫          a              ∞            n    2    x      e          −              n        2                    x        2                d  x  =      1    2    exp  ⁡  (  −      n    2        a    2    )  .

Show that <mstyle displaystyle="true" scriptlevel="0"> <munder> <mo movablelimits="t

Answered question

Answer & Explanation

New Questions in Pre-Algebra