Evaluating lim x → 1 ∫ 1 x ln 2 ⁡ ( t ) d t ( x − 1 ) 3

Question

Evaluating       lim          x      →      1                          ∫                  1                          x                            ln        2            ⁡      (      t      )      d      t              (      x      −      1              )        3

verzaadtwr · Accepted Answer

In terms of the evaluation of the limit: applying L&#039;Hôpital&#039;s rule once forms      lim          x      →      1                          ∫                  1                          x                            ln        2            ⁡      (      t      )            d      t              (      x      −      1              )        3              =      lim          x      →      1                          ln        2            ⁡      (      x      )              3      (      x      −      1              )        2              ,where the chain rule implies that the derivative of the denominator is      d          d      x            [    (    x    −    1          )      3        ]    =  3  (  x  −  1      )    2    .From here you can apply L&#039;Hôpital&#039;s rule two more times to find      lim          x      →      1                          ln        2            ⁡      (      x      )              3      (      x      −      1              )        2              =      lim          x      →      1                  ln      ⁡      (      x      )              3      x      (      x      −      1      )        =      lim          x      →      1            1          x      (      6      x      −      3      )        =      1    3    .To answer your second question, the result follows from the fundamental theorem of calculus. Let f  be a continuous, real-valued function defined on the closed interval [a,b]. Suppose F is defined for all   x  ∈  [  a  ,  b  ] by  F  (  x  )  =      ∫    a          g      (      x      )        f  (  t  )    d  t  .Then      F    ′    (  x  )  =      d          d      x            ∫          a              g      (      x      )        f  (  t  )    d  t  =  f  (  g  (  x  )  )      g    ′    (  x  )  .Note that we could change the lower integration constant to any other positive real constant   a  &amp;gt;  0 (as the fundamental theorem of calculus would still hold) to deduce      F    ′    (  x  )  =      d          d      x            ∫          a              x            ln    2    ⁡  (  t  )    d  t  =      ln    2    ⁡  (  x  )  .However, if the upper limit of integration was       x    2   instead of x, then we would find      F    ′    (  x  )  =      d          d      x            ∫          a                      x        2                  ln    2    ⁡  (  t  )    d  t  =  2  x      ln    2    ⁡  (      x    2    )  .

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