       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:=
f[x_] = (a^2 - a*x)/(a - Sqrt[a*x]) /. {a -> -1}

Out=
1 + x
-------------
-1 - Sqrt[-x]

You can easily check (e.g. by using Table) that Mathematica's answer 0 is

Of course if you put a=1 the correct answe will be indeed your answer as you
can check will Mathematica:

In:=
Limit[(1 - x)/(1 - Sqrt[x]), x -> 1]

Out=
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=
2
a + Sqrt[a ]

In:=
Simplify[%, a > 0]

Out=
2 a

In:=
Simplify[%%, a < 0]

Out=
0

--
Andrzej Kozlowski
Toyama International University, JAPAN

For Mathematica related links and resources try:
<http://www.sstreams.com/Mathematica/>

```

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