问题及解答

利用极限的定义证明

(1)  $\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{x+1}{2x}}={\displaystyle \frac{1}{2}}$

(2)   $\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{x^2-1}{x-1}}=2$

(2)

$\mathrm{\forall }\epsilon >0$, $\mathrm{\exists }\delta =\epsilon$, 当 $0<|x-1|<\delta$ 时,

$$|\frac{x^2-1}{x-1}-2|=|x+1-2|=|x-1|<\delta =\epsilon ,$$

故 $\underset{x\to 1}{lim}{\displaystyle \frac{x^2-1}{x-1}}=2$.