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MathGroup Archive 2001

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Re: "Limit involving square root"

  • To: mathgroup at smc.vnet.net
  • Subject: [mg30291] Re: "Limit involving square root"
  • From: scundal at yahoo.com (Hein H)
  • Date: Sun, 5 Aug 2001 16:18:35 -0400 (EDT)
  • References: <9kg0ii$idk$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

George Woodrow III <georgevw3 at mac.com> wrote in message news:<9kg0ii$idk$1 at smc.vnet.net>...
> I tried plotting the original function from 0 to 10^16, and got a lot of 
> artifacts. There is a strong horizontal line at 5, and spikes. What you 
> got were probably artifacts.
> 
> If you rewrite the function as x - Sqrt[(x - 9) (x - 1)], it should be 
> clear that as x approaches infinity, this approximates x - Sqrt[x^2] = 0.
> 
> Don't know why Mathematica is confused. Nlimit[] reports 5 for as the limit and 
> 6 for the function + 1.

   The actual limit is 5.  Try plotting the function from 10 to 1000. 
Here is the reasoning for a limit of 5:

For x > 10,

Let f[x] = x - Sqrt[(x - 9) (x - 1)]


  (x - Sqrt[(x - 9) (x - 1)]) (x + Sqrt[(x - 9) (x - 1)])
= -------------------------------------------------------
                 (x + Sqrt[(x - 9) (x - 1)])


        x^2 - (x-9) (x-1)
= -------------------------------
     (x + Sqrt[(x - 9) (x - 1)])

       10 x - 9
= -------------------------
  (x + Sqrt[(x - 9) (x - 1)])



Now using L'Hospitals rule,

Limit[f[x]] = 10/Limit[ 1 + x/ Sqrt[ (x-9) (x-1) ]] = 5.


Cheers,
Hein H


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