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Re: l'Hopital's Rule

• To: mathgroup at smc.vnet.net
• Subject: [mg24418] Re: l'Hopital's Rule
• From: "John L. Jechura Jr." <JLJechura at MarathonOil.com>
• Date: Tue, 18 Jul 2000 00:58:15 -0400 (EDT)
• Organization: Marathon Oil Company
• References: <8k3oi3$42k@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

There are actually two answers depending upon the restrictions on a. If a>0
(as you have assumed), then the limit is zero. However, if a<0, then
a-Sqrt[a^2] is not zero but rather 2a; in this case, the limit would be
zero.

John Jechura
Sr. Engineer
Marathon Oil Company

<heathw at in-tch.com> wrote in message news:8k3oi3$42k at smc.vnet.net...
> 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
>
>