1. x_n is not a Cauchy sequence means:
there is some e>0, such that no matter which positive integer N you choose, there is some n and m at least N such that |x_n-x_m|>=e.
With x_n=n choose e=1/2, n=N+1, and m=N+3, so that |x_n-x_m|=2>1/2.
2. Given e>0 ,we must find d>0 such that d>|x|, x not 0, implies e>|xf(x)|.
Since lim_{x->0} f(x)=L, choose d1 such that d1>|x|, x not 0, implies e1>|f(x)-L|. Thus for these x's, L+e1>f(x)>L-e1, so |f(x)| is bounded on this set, say by M>0.
For d1>|x|, x not 0, M|x|>=|xf(x)|. Now choose d=min{d1,e/M}.
3. (a) Diverges: Proof #1: Since lim_{n->infty} (1+1/n)^n=e, the root test gives lim_{n->infty} ((n+1)/n)^n*2^(-1)=e/2>1.
Proof #2: (1+1/n)^(n^2)= ((1+1/n)^n)^n>=(1+1)^n by the binomial thoerem, so the nth term is bounded below by 1, and does not go to 0.
(b) Converges. The Nth partial sum of the b_n, is the 2Nth partial sum of the a_n, starting at n=1.
(c) (Hardest problem on the exam.) Either could occur. If b_n=a_n we have convergence. Here is an example with divergence. Let a_n=(-1)^n/n, the alternating harmonic, which converges. Let b_n=(-1)^n/n+1/nlog(n). b_n diverges since \sum_{n} 1/nlog(n) diverges.
4. The groph of f(x) is a parabola, vertex at (0,0) pointing up to (1/2,1/2), and another parabola, vertex (1,1), pointing down to (1/2,1/2). the first parabola lies below the line y=x on (0,1/2) , since x>=2x^2 there, while the second parabola lies above the line y=x on (1/2,1). Thus x_n is decreasing if 1/2>a>0, and increasing if 1>a>1/2. If a=0, x_n=0, if a=1, x_n=1, and if a=1/2, x_n=1/2. Thus x_n is always a monotone bounded sequence, so it converges.Solving 2x^2=x, and 1-2(x-1)^2=x gives g(a)=0, if a \in [0,1/2), g(1/2)=1/2, gives g(a)=1, if a \in (1/2,1].