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Real Analysis




                    Notes
                                                                    3 éæ  7 ö æ  11ö æ  5 ù ö
                                                                   x ç ê  2 + ÷ ç 3 -  ÷ ç  4 + ÷ ú
                                                                     ë è  x ø è  x ø è  x ø û
                                                              = lim
                                                                x      3 æ  1  1 ö
                                                                        x ç  4 +  -  3 ÷
                                                                          è  x  2  x ø
                                   We divide the numerator and denominator by x  since x  is neither zero nor .
                                                                          3
                                                                                3
                                                                   (2x 7)(3x 11)(4x 5)
                                                                                  +
                                                                            -
                                                                      +
                                                              = lim       3
                                                                x     4x +  x -  1
                                                                   æ   7 ö æ  11ö æ  5 ö
                                                                   ç 2 + ÷ ç 3 -  ÷ ç 4 + ÷
                                                                                         ´
                                                                                           ´
                                                                   è   x ø è  x ø è  x ø  2 3 4
                                                              = lim                   -       = .6.
                                                                x        1   1          4
                                                                        4 +  -
                                                                           x  2  x 3
                                                            2
                                                           x -  9
                                          Example: Find  lim
                                                          2
                                                      x 3 x -  4x +  3
                                                        2
                                                                     -
                                                                          +
                                                       x -  9      (x 3)(x 3)
                                   Solution:      lim  2      = lim
                                                  x 3 x -  4x +  3  x 3 (x 3)(x 1)-  -
                                                       2
                                                                    +
                                                      x -  9       x 3
                                   Hence         lim  2       = lim
                                                 x 3 x -  4x +  3  x 3 x 1-
                                                                     +
                                                                lim  (x 3)
                                                                x  3     6
                                                              =         =   = 3.
                                                                     -
                                                                lim  (x 1)  2
                                                                x  3
                                                     2
                                                    x -  9
                                   The function f(x) =      is not defined at x = 3. But we are considering only the values of
                                                   2
                                                  x -  4x +  3
                                   the function at those points x in a neighbourhood of 3 for which x  3 and hence we can cancel
                                   x – 3 factor.
                                                            (1 x) 1 2  -  1
                                                              +
                                          Example: Evaluate  lim  1 3  .
                                                         x 0 (1 x)+  -  1
                                   Solution: To make the problem easier, we make a substitution which enables us to get rid of
                                   fractional powers 1/2 and 1/3. L.C.M. of 2 and 3 is 6. So, we put 1 + x = y .
                                                                                             6
                                   Then we have
                                                                                   2
                                                                    3
                                                                               -
                                                      +
                                                    (1 x) 1 2  -  1  y -  1  (y 1)(y +  y 1)
                                                                                       +
                                                lim     1 3   = lim  2  = lim
                                                                                -
                                                                                     +
                                                 x 0 (1 x)+  -  1  y 1 y -  1  y  1  (y 1)(y 1)
                                                                    2
                                                                        +
                                                                   y +  y 1  3
                                                              = lim        -  .
                                                                      +
                                                                y  1  y 1  2
                                   Self Assessment
                                   Fill in the blanks:
                                   1.  The intuitive idea of limit was used both by Newton and Leibnitz in their independent
                                       invention of Differential Calculus around ........................
                                   2.  The limit of a function that a point a is meaningful only if a is a limit point of its ..................
                                                                        2
                                   3.  For a function f: R  R defined by f(x) = x , find its limit when x tends to 1 by the .
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