Path: ...!weretis.net!feeder8.news.weretis.net!feeder1-2.proxad.net!proxad.net!feeder1-1.proxad.net!cleanfeed1-b.proxad.net!nnrp4-1.free.fr!not-for-mail
Date: Sat, 13 Jan 2024 12:33:20 +0100
MIME-Version: 1.0
User-Agent: Mozilla Thunderbird
Subject: Re: Limite
Newsgroups: fr.sci.maths
References: <ai5qMT9SfdsE8vSSatTKMJjt4Bk@jntp> <untpdd$3tac1$2@dont-email.me>
Content-Language: fr
From: Samuel Devulder <samuel.devulder@laposte.net.inalid>
In-Reply-To: <untpdd$3tac1$2@dont-email.me>
Content-Type: text/plain; charset=UTF-8; format=flowed
Content-Transfer-Encoding: 8bit
X-Antivirus: Avast (VPS 240113-0, 13/1/2024), Outbound message
X-Antivirus-Status: Clean
Lines: 19
Message-ID: <65a27500$0$8239$426a74cc@news.free.fr>
Organization: Guest of ProXad - France
NNTP-Posting-Date: 13 Jan 2024 12:33:20 CET
NNTP-Posting-Host: 88.167.72.245
X-Trace: 1705145600 news-2.free.fr 8239 88.167.72.245:28937
X-Complaints-To: abuse@proxad.net
Bytes: 1593

Le 13/01/2024 à 11:41, efji a écrit :

>> stipule que : lim{x->e} (x-e)/(ln(x)-1) = e
>> Comment ça se démontre ?
> 
> x = e+y, y petit. (|y|<<1).
> ln(x) = ln(e+y) = ln(e) + y/e + o(y) = 1 + y/e + o(y)
> (x-e)/(ln(x)-1) = y/(y/e+o(y)) = e + o(y) -> e lorsque y->0

On peut aussi utiliser la règle de l'Hôpital:

On note f(x)=(x-e), g(x) = ln(x)-1, f et g tendent toutes deux 0 quand 
x->e. Donc lim f(x)/g(x) = lim f'(x)/g'(x) quand x->e

Or f'(x) = 1 et g'(x)=1/x, soit f'(x)/g'(x) = x. On a donc:

	lim f(x)/g(x) = lim f'(x)/g'(x) = lim x = e lorsque x->e

sam.