a) |x - 2| <1 ----> --------- < 2-x/2x < ----------

b) |x| < 1 ------> ---------- < x^2 +2x < -----------

c) |x + 1| < 1 ------> ----------- < x^2 - 1 < ---------

d) |x - 2| < 1 ------> ------- < x^2 +2x+ 4 < -------

delta- epsilon proofs:

(I know that when finding delta we need to use the equation |f(x) - L| < epsilon and then substitute delta for epsilon but I am confused as to how to use the bounds from the previous problem in this question. )

a) as x approaches 2, the limit of x^2 = 4

b) as x approaches, the limit of x^ -2 = 4

c) as x approaches 2, the limit of 1/x = 1/2

d) as x approaches 2, the limit of x^3 = 8