Strona 1 z 1
granice
: 20 sty 2007, o 13:49
autor: agataga1
Mam jeszcze problem z takimi granicami:
\(\displaystyle{ \lim_{x\to1}\frac{e^{x} - e}{x - 1}}\)
\(\displaystyle{ \lim_{x\to0}\frac{e^{sin x} - 1}{x}}\)
granice
: 20 sty 2007, o 14:37
autor: Lorek
\(\displaystyle{ \frac{e^x-e}{x-1}=\frac{e(e^{x-1}-1)}{x-1}}\)
Podstawienie
\(\displaystyle{ x-1=t,\; t\to 0}\)
\(\displaystyle{ \lim\limits_{x\to 1}\frac{e(e^{x-1}-1)}{x-1}=\lim\limits_{t\to 0}\frac{e(e^t-1)}{t}=e}\)
[ Dodano: Sob Sty 20, 2007 2:39 pm ]
\(\displaystyle{ \lim\limits_{x\to 0}\frac{e^{\sin x}-1}{x}=\lim\limits_{x\to 0}\frac{e^{\sin x}-1}{\sin x} \frac{\sin x}{x}=1}\)