1. Let x 1=x2=0, and get f (0) = f (0) 2; Because f(0) is not zero, f (0) =1;

, defined by continuous limit, namely lim (△ x→ 0) △ y = 0;

Let x be any point on r, and there is an increment △ x at x; therefore

lim(△x→0)△y = lim(△x→0)[f(x+△x)-f(x)]

= lim(△x→0)[f(x)f(△x)-f(x)]= lim(△x→0)[f(x)(f(△x)- 1)]

That is, lim(△x→0)△y=0, so F is continuous at X. Because of the arbitrariness of X, F is continuous everywhere.

2 obviously xn>0; xn+ 1 = 1/2(xn+a/xn)>:= 1/2 * 2√a =√a；

That is, xn> = √ a;

So xn+1/xn =1/2 (1+a/xn2) < =1(because xn >；; =√a)

That is xn+ 1 < Xn.

To sum up, Xn is a monotonically decreasing sequence with a lower bound, so its limit exists and is set to L.

therefore

Xn+ 1= 1/2(Xn + a/Xn)

Take the limit of the above formula

L = 1/2(L+a/L)；

Get L=√a or -√a (rounded).