Evaluate:

               (1+ x)^1/x – e + e*x/2
   Limit     --------------------------
   x → 0             x²

Note: e denotes the Euler’s number.

Let                y = (1 + x)^1/x

then         ln(y) = (1/x)ln(1 + x)      now, using the power series expansion..
                        = (1/x)[x – x²/2 + x³/3 – x⁴/4 + …]
                        = 1 – x/2 + x²/3 – x³/4 + …

which gives    y = e^[1 – x/2 + x²/3 – x³/4 + …]
                        = [e^1][e^(-x/2)][e^(x²/3)][e^(-x³/4)]….

Now using the expansion: e^t = 1 + t + t²/2! + t³/3! … in each bracket.

                    y  = e[1 + (-x/2) + (-x/2)²/2! + …][1 + x²/3 + …][1 + …]….
                        = e[1 –x/2 + x²/3 + x²/8 + terms involving x³ and higher powers]
                        = e[1 – x/2 + 11x²/24]  +  O(x³)

Substituting this into the formula given:

[(1 + x)^(1/x) – e + ex/2]/x²   = e[1 – x/2 + 11x²/24] –e + ex/2]/x² + O(x)
                                              = 11e/24  +  O(x)

So,  in the limit as x tends to 0, the value will be   11e/24 ( = 1.24587..)

Edited on November 10, 2009, 11:18 pm

Posted by Harry on 2009-11-10 23:14:24