Using L'Hopital's Rule

Date: 01/22/99 at 05:17:12
From: Mirza
Subject: What is the limit of ....

Could you give me the proof that e^(-1/3) is the limit for

   y = (arctan(x)/x)^(1/(x^2)) 

when x -> 0?

Date: 01/22/99 at 08:48:31
From: Doctor Anthony
Subject: Re: What is the limit of ....


Take logs:

   ln(y) = (1/x^2)ln[tan^(-1)(x)/x]

           ln[tan^(-1)(x)/x]
   ln(y) = -----------------
                 x^2   

Note that tan^(-1)(x)/x  -> 1 as x -> 0 and ln(1) = 0. So the righthand 
side tends to the value 0/0 as x -> 0, and we can apply l'Hopital's 
rule. Differentiating the top and bottom lines separately we have:

   [x/tan^(-1)(x)][x/(1+x^2)-tan^(-1)(x)]/x^2 
   ---------------------------------------------
                      2x

      x - tan^(-1)(x) (1+x^2)
    = ------------------------   
      2x^2 tan^(-1)(x) (1+x^2)

As x -> 0, this is still 0/0, so we repeat the process. Differentiating 
the top and bottom separately

      1 - [(1+x^2)/(1+x^2) + tan^(-1)(x) 2x]
   --------------------------------------------
   2(x^2+x^4)/(1+x^2) + 2 tan^(-1)(x) (2x+4x^3)

which simplifies to

        -tan^(-1)(x)
   -------------------------  
    x + 2(1+2x^2)tan^(-1)(x)  

This is still 0/0 so repeat the process. Differentiate:

               -1/(1+x^2)
   --------------------------------------   
   1 + 2(1+2x^2)/(1+x^2) + 8x tan^(-1)(x)

Then multiply top and bottom by (1+x^2):

                   -1
  ------------------------------------------
  1 + x^2 + 2(1+2x^2) + 8x(1+x^2)tan^(-1)(x)

Putting x = 0, this reduces to:

    -1      -1
   ----- =  ---
   1 + 2     3
                     
So the limit of ln(y) = -1/3. Thus the limit of y = e^(-1/3).

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/