106 - 2a  Look at |(n+1)/n - (m+1)/m| = |(mn+m-mn-n)/mn| LT 1/n + 1/m; 
if m LT n this is less than 2/m; let H(epsilon) = 2/epsilon.

(I misread the assignment, but thought I should leave this)
*      6   Let S be the supremum of the sequence, by definition of supremum, 
*there is a point within epsilon/2 of the supremum, by definition of monotone
*all subsequent members of the sequence are within epsilon/2 of the supremum,
*by the triangle inequality any two subsequent members are within epsilon of
*each other.

      5  Choose epsilon equal .5, beyond H(epsilon) all members of the sequence
differ by at most one half: since they are integers, they are equal.

      9  We know Sum(r^n) = 1/(1-r) (the sum starts at n=0), hence
Sum_(n=k to infinity)(r^n) = (r^k)/(1-r).  We can choose k = H(epsilon) so
this is less than epsilon.

109 - 2a  use (1,2,3,4,5, ...) for both
       b  use (x_n)=(n^2), (y_n)=(n)

      3   Given epsilon which you want x_n to be less than, use K(1/epsilon)
which assures 1/(x_n) is large; given alpha which you want 1/(x_n) to be larger
than, use K(1/alpha) which assures x_n is small.

      8a  (n^2 + 2)^.5 GT n, hence properly diverges
       b  n^.5 / (n^2 + 1) LT 1/n, hence converges to 0
       c  (n^2 + 1)^.5 / n^.5 GT n^.5, hence properly diverges
       d  sin(n^.5) is bounded, but will get arbitrarily close to +1 and -1
infinitely often, so it diverges.

119 - 1c  want |(x-1)(x+1)| LT 1/n  make delta LT 1 so x+1 LT 2; make delta  LT
1/2n so (1-x)(1+x) LT (1/2n)(2) = 1/n  delta = min (1, 1/2n)

      6   |x^2 - c^2| = |(x-c)(x+c)| LT |x-c|(a+a) [x,c LT a] therfore, let
delta equal epsilon/2a so the product will be less than epsilon when
|x-c| LT delta

     10b  |x/(1+x) - 1/2| = |(2x-1-x)/(2(1+x))|  letting delta less than 1/2
the denominator is greater than 1, hence the product is less than |x-1|
let delta = min (.5, epsilon) and the product will be less than epsilon.
I am confused by the sequential criterion as a seperate demonstration, any
sequence which converges to 1 will get within a delta neighborhood of 1 for all
delta, and if delta is less than .5, the function will be within delta of 1/2
hence the sequence must converge to .5.