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Re: l'Hopital's Rule
*To*: mathgroup at smc.vnet.net
*Subject*: [mg24314] Re: [mg24309] l'Hopital's Rule
*From*: Andrzej Kozlowski <andrzej at tuins.ac.jp>
*Date*: Sun, 9 Jul 2000 04:52:27 -0400 (EDT)
*Sender*: owner-wri-mathgroup at wolfram.com
on 00.7.7 1:11 PM, heathw at in-tch.com at heathw at in-tch.com wrote:
> Hi,
> When I input this:
> Limit[(a^2 - a*x)/(a - Sqrt[a*x]), x -> a]
> The output is:
> 0
> The output should be 2*a.
> Can't Mathematica 4 use l'Hopital's Rule?
> Thanks,
> Heath
>
>
Actually, you are at least as wrong as Mathematica here. Try, for example,
a=-1.
In[1]:=
f[x_] = (a^2 - a*x)/(a - Sqrt[a*x]) /. {a -> -1}
Out[1]=
1 + x
-------------
-1 - Sqrt[-x]
You can easily check (e.g. by using Table) that Mathematica's answer 0 is
now correct, and your answer -2 is wrong.
Of course if you put a=1 the correct answe will be indeed your answer as you
can check will Mathematica:
In[7]:=
Limit[(1 - x)/(1 - Sqrt[x]), x -> 1]
Out[7]=
2
In fact Mathematica can do the general case, provided you first tell it to
simplify the input:
Limit[FullSimplify[(a^2 - a*x)/(a - Sqrt[a*x])], x -> a]
Out[39]=
2
a + Sqrt[a ]
In[10]:=
Simplify[%, a > 0]
Out[10]=
2 a
In[11]:=
Simplify[%%, a < 0]
Out[11]=
0
--
Andrzej Kozlowski
Toyama International University, JAPAN
For Mathematica related links and resources try:
<http://www.sstreams.com/Mathematica/>
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