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Basic Mathematics – I




                    Notes                                                                [Dividing throughout by x]

                                                             Ê   x  x 2     ˆ
                                                            x 1+  Á  +  + .......... ˜
                                                         =   Ë   2!  3!     ¯
                                                                    x

                                                         =   Ê 1+  x  +  x 2  ˆ
                                                            Á Ë  2!  3!  + .......... ˜ ¯

                                                                           ˆ
                                                            Ê
                                                    x
                                                   e - 1 - 1
                                                                    2
                                   \            limlim  e x   =   lim +  Ê Á  x x x x  2 .................... ˆ ˜
                                                         lim 1=
                                                                    +
                                                                     + +
                                                        =
                                                                        +
                                                                  + 1
                                                                           ˜
                                                            Á
                                                x Æ  0 x Æ  0 x x  x Æ  0  Ë xÆ  0  Ë  ! 2 2  ! 3 !  ! 3  ¯  ¯
                                                         =  1 + 0 + 0 + ...... = 1
                                                    x
                                   Thus,        lim  e - 1  =  1
                                                xÆ0  x
                                          Example 6:
                                   Examine the behaviour of the function in each of the following:
                                   (i)            when x → 2, x → – ∞ and x → ∞
                                   (ii)            when x → 1, x → – ∞ and x → ∞


                                   (iii)             when x → + ∞


                                   Show the behaviour by sketching graph, indicating the asymptotes of the function.
                                   Solution:
                                   Note that y is not defined in each of the above cases.


                                   (i)  LHL


                                       RHL




                                       Also




                                       and


                                       Note that there is a vertical asymptote at x = 2 and a horizontal asymptote    to the
                                       function.


                                       The behaviour of the function is shown in Fig. 7.3.




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