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    UAE_Math_Grade_12_Vol_1_SE_718383_ch3
                                      GO01962-Smith-v1.cls
                                                         July 4, 2016
                                              Rather than apply l’Ĥopital’s Rule to this last expression, we first simplify the
                                          expression, by multiplying top and bottom by (x + 1). We now have
                                                                              1 −  1
                                                    [            ]                x + 1      (     )
                                                 lim     1    −  1                            x + 1
                                                 x→0 ln (x + 1)  x  = lim (  1  )             x + 1
                                                                    x→0
                                                                          x + 1  x + ln (x + 1)(1)
                                                                            (x + 1) − 1   ( )
                                                                                            0
                                                                  = lim
                                                                    x→0 x + (x + 1) ln (x + 1)  0
                                                                                d  (x)
                                                                                dx
                                                                  = lim                       By l’Ĥopital’s Rule.
                                                                    x→0 d  [x + (x + 1) ln (x + 1)]
                                                                        dx
                                                                                     1               1
                                                                  = lim                            =   ,
                                                                    x→0                        1     2
                                                                        1 + (1) ln (x + 1) + (x + 1)
                                                                                             (x + 1)
                                          which is consistent with Figure 4.19.


                y                         EXAMPLE 2.7     The Indeterminate Form 0 ⋅ ∞
                                                     ( 1   )
             0.4                          Evaluate lim  ln x .
                                                  x→∞ x
             0.3
                                          Solution This limit has the indeterminate form (0 ⋅ ∞). From the graph in
                                          Figure 4.20, it appears that the function is decreasing very slowly toward 0 as
             0.2                          x → ∞. It’s easy to rewrite this in the form  ∞  and then apply l’Ĥopital’s Rule.
                                          Note that                           ∞
             0.1                                               (  1  )      ln x  ( ∞  )
                                                           lim    ln x = lim
                                                           x→∞ x        x→∞ x      ∞
                                       x
                   20  40  60  80  100                                       d  ln x
                                                                             dx
                     FIGURE 4.20                                      = lim   d     By l’Ĥopital’s Rule.
                                                                        x→∞
                       y =  1 ln x                                            dx  x
                          x
                                                                             1
                                                                             x   0
                                                                      = lim    =   = 0.
                                                                        x→∞ 1    1
                                                                                                    ∞
                                                                                            0
                                                                                               0
                                          Note: If lim[f(x)] g(x)  has one of the indeterminate forms 0 , ∞ or 1 , then, letting
                                                 x→c
                                          y = [f(x)] g(x) ,wehavefor f(x) > 0 that
                                                                 ln y = ln[f(x)] g(x)  = g(x) ln [f(x)],
                y
                                          so that lim ln y = lim{g(x) ln [f(x)]} will have the indeterminate form 0 ⋅ ∞, which we can
                                                x→c     x→c
              20                          deal with as in example 2.7.
              15
                                          EXAMPLE 2.8     The Indeterminate Form 1  ∞
              10                                       1
                                          Evaluate lim x x−1 .
                                                  x→1 +                                          ∞
         Copyright © McGraw-Hill Education   0.5 FIGURE 4.21  2.0  x  graph in Figure 4.21, it appears that the limit is somewhere around 3. We define
              5
                                          Solution First, note that this limit has the indeterminate form (1 ). From the
                                               1
                         1.0
                              1.5
                                          y = x x−1 , so that
                                                                                 1
                                                                           1
                                                                  ln y = ln x x−1 =
                                                                                    ln x.
                            1
                                                                                x − 1
                       y = x x−1
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