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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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