Exercise 1.6. Alyssa P. Hacker doesn't see why if needs to be provided as a special form. ``Why
can't I just define it as an ordinary procedure in terms of cond?'' she asks. Alyssa's friend Eva Lu
Ator claims this can indeed be done, and she defines a new version of if:
(define (new-if predicate then-clause else-clause)
(cond (predicate then-clause)
(else else-clause)))
Eva demonstrates the program for Alyssa:
(new-if (= 2 3) 0 5)
5
(new-if (= 1 1) 0 5)
0
Delighted, Alyssa uses new-if to rewrite the square-root program:

(define (sqrt-iter guess x)
(new-if (good-enough? guess x)
guess
(sqrt-iter (improve guess x)
x)))
What happens when Alyssa attempts to use this to compute square roots? Explain.

answer: 
1.Normal Order 
  In this order, it will fully expand  and won't be evaluate until it is necessary. So it won't evaluate the (sqrt-iter (improve guess x) x)) which is the new-if's argument. , it will get the right answer.
  I have tested it in Racket with "Lazy Racket" language, and it works as expect.


2.Applicative Order
  In this order, the interpreter will evaluate sqrt-iter->new-if, the (sqrt-iter (improve guess x) x)) will evaluate, and we again call sqrt-iter->new-if and the (sqrt-iter (improve guess x) x)) will evaluate again , so it is a infinite loop.