Suppose I have the limit: lim t → 0 + ∫ 0 t + 4 1 + t 4 x d x

Question

Suppose I have the limit:      lim          t      →              0        +                  ∫    0                            t                +        4                  1    +          t              4                  x                        d  x

laure6237ma · Accepted Answer

To prove the limit is 2 we must show that for any   ϵ  &amp;gt;  0 there exists   δ  &amp;gt;  0 such that if   0  &amp;lt;  t  &amp;lt;  δ, then                    (*)                              |                      ∫            0                                                            t                                +                4                                                          1            +                          t                              4                                  x                                                                        d          x          −          2          |                &amp;lt;        ϵ            Note that      |          ∫      0                                    t                    +          4                            1      +              t                  4                      x                                    d    x    −    2    |    ⩽      |          ∫      2                                    t                    +          4                            1      +              t                  4                      x                                    d    x    +          ∫      0              2                    1      +              t                  4                      x                                    d    x    −    2    |      ⩽                              |                      ∫            2                                                            t                                +                4                                                          1            +                          t                              4                                  x                                                                        d          x          |                ⏟                    A        +                              |                      ∫            0                          2                                            1            +                          t                              4                                  x                                                                        d          x          −          2          |                ⏟                    B      For the term A, with   x  ⩾  2 we have  A  =      |          ∫      2                                    t                    +          4                            1      +              t                  4                      x                                    d    x    |    ⩽      1    +          t              4                  2                          (                  t            +      4        −    2    )  Since the RHS depends continuously on t there exists       δ    1    &amp;gt;  0, then   A  &amp;lt;  ϵ      /    2Term B is easily handled by showing that the improper integral converges uniformly for t in any bounded interval   [  0  ,  C  ]. Since       |          1      +              t                  4                      x                                |    ⩽      1    +          C              4                  x                     it follows by the Weierstrass M-test that the improper integral appearing in B is uniformly convergent and, thus,      lim          t      →      0      +            ∫    0          2            1    +          t              4                  x                        d  x  =      ∫    0    2        lim          t      →      0      +            1    +          t              4                  x                        d  x  =  2Hence, there exists       δ    2    &amp;gt;  0 such that when   0  &amp;lt;  t  &amp;lt;      δ    2   we have  B  =      |          ∫      0              2                    1      +              t                  4                      x                                    d    x    −    2    |    &amp;lt;      ϵ    2  Therefore, we conclude the limit is 2 since if   0  &amp;lt;  t  &amp;lt;  δ  =  min  (      δ    1    ,      δ    2    ), then      |          ∫      0                                    t                    +          4                            1      +              t                  4                      x                                    d    x    −    2    |    ⩽  A  +  B  &amp;lt;      ϵ    2    +      ϵ    2    =  ϵ  ,

Petrovcic2x · Accepted Answer

Use the substitution       u    2    =      t    +  4 so that   u  →      2    +   and the integral is reduced to      ∫    0    u        1    +                  (                  u          2                −        4                  )          2                            4                  x                        d  xLet the integrand be denoted by   f  (  u  ,  x  ) then we have  0  ≤  f  (  u  ,  x  )  −  1  =            (              u        2            −      4      )              4              x              ⋅            1              1        +                  1          +                                                    (                                  u                  2                                −                4                                  )                  2                                                            4                                  x                                                                          ≤            (              u        2            −      4              )        2                    8              x            Thus we have  1  ≤  f  (  u  ,  x  )  ≤  1  +            (              u        2            −      4              )        2                    8              x            Integrating with respect to x on   [  0  ,  u  ] we get  u  ≤      ∫    0    u    f  (  u  ,  x  )    d  x  ≤  u  +            (              u        2            −      4              )        2                    u              4  Letting   u  →      2    +   we get desired limit as 2

Suppose I have the limit: <munder> <mo movablelimits="true" form="prefix">lim <mrow clas

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New Questions in Calculus 1