       Re: How do I solve limits in Mathematica?

• To: mathgroup at smc.vnet.net
• Subject: [mg17192] Re: How do I solve limits in Mathematica?
• From: "Allan Hayes" <hay at haystack.demon.co.uk>
• Date: Tue, 20 Apr 1999 01:21:03 -0400
• References: <7f1esn\$krh@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```Jeffrey:

Limit[2x-1, x->0]

-1

Limit[2x^2+1, x->1]

3

But that tells us little about what "limit" means.

Let's look at  Limit[2x-1, x->a] (your example is with a =0)
Put x = a+h and expand

Expand[(2x-1)/.x:>a+h]

-1+2 a+2 h

Now we can see that as h ->0 (that is as x->a) 2x-1 -> -1 +2a
Similarly

Expand[(2x^2+1)/.x->a+h]

1 + 2a^2 + 4a h + 2h^2

Best wishes

---------------------
Allan Hayes
Mathematica Training and Consulting
Leicester UK
www.haystack.demon.co.uk
hay at haystack.demon.co.uk
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565

Jeffrey A. Soesbergen <Voyager_2 at cable.A2000.nl> wrote in message
news:7f1esn\$krh at smc.vnet.net...
> How can I solve the following expressions in Mathematica? Anyone?
>
> 1)
>
> lim (2x-1) == -1
> (x->0)
>
> 2)
>
> lim (2x^2+1) == 3
> (x->1)
>
> I can't figure out how to solve those expressions, so if anyone can tell
me...
> thanks !
>
> J.
>

```

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