151 - 1c The numerator is continuous because sin(x) is cont., the |.| of a 
cont. function is cont., the sum of cont fntns is cont, the sqrt of a positive 
cont function is cont.  The ratio is continuous except where x=0, because ratios 
of continuous functions are continuous except where the denominator is 0.

      8  If x is irrational, there is a sequence of real numbers (x_n) which 
approaches it.  Since f(x_n)=g(x_n) for all n, they have the same limit (which 
exists since f and g are continuous) and that limit is f(x)=g(x) because f 
and g are continuous.

      15 If f(x) GT g(x), .5(f+g) + .5|f-g| = f, if f LT g, .5(f+g) + .5|f-g| 
= g, hence h is as given.  Since the sum, difference, and absolute value of 
continuous functions is continuous, h is continuous.

159 - 1  Because f is cont, it obtains its minimum on a closed bounded interval
(Thm 5.3.4).  That minimum is GT 0, because all values of f are on that
interval. Let alpha = half the minimum.

      8  By continuity, f(w) is LT or equal to 0. (If w in W, f(w) LT 0, if
w not in W, there is a sequence in W which approaches it, and f(w) LT or 
equal to 0 by continuity.)  If f(w) LT 0, let f(w) = y LT 0.  The IVT implies 
there exists an x between w and b with f(x) = y/2.  This contradicts that w
is the sup of W. 

      12  A open [respectively closed] interval with both endpoints positive
will be mapped to an open [respectively closed] interval since the endpoints
which are not [respectively are] included are mapped to the enpoints which are
not [respectively are] included.  The same argument holds if both endpoints
are negative.  (Or if one enpoint is zero.)  If the enpoints span 0, 0 will be
an included endpoint in the image, and the inclusion of the endpoint which is
greatest in magnitude will determine the inclusion of the image endpoint.

171 - 1  (Look at |1/x - 1/c| = |(c-x)/xc|.  If delta = min(.5, epsilon/2),
xc GT .5, hence |(c-x)/xc| LT epsilon if |x-c| LT delta.

      10 Let M be an upper bound for A and m a lower bound.  Let epsilon 
equal 1.  Let delta be the associated delta.  Then there are (M-m)/delta
intervals on which f varies by 1 or less.  One can choose any point in each 
interval where f is defined, and evalute f there, then take the supremum and 
infinum of those evaluations (which will be among the evaluations, since there 
is a finite number).  1 less than the imfinum and 1 more than the supremum
provide bounds on f.  [N.B., I was worried about gaps in A where f is not 
defined which would preclude asserting f varies by less than (M-m)/delta
over A.]

      12 If f is continuous on [0, infty), it is uniformly continuous on [0,2a].
For any epsilon, choose delta = min (.5a, delta(for [0,2a]), 
delta(for [a, infty) )  ).  That delta demonstrates uniform continuity, since
it works on [0, infty).