202 METHODS OF MATHEMATICAL PROOF, PART II Chapter 6- Use an epsilon-delta argument to prove:
(a) lim,,,(-3x + 6) = -6 (b) lim,,, x2 + 5 = 86
(c) lirn,,, 2x2 + 3x +^1 =^6 - Use an epsilon-delta argument to prove:
(a) li,f(x)=ll, where f(x)=
2, x=3
x, x rational
*(b) lirn,,, g(x) = 0, where g(x) =
0, x irrational 1
(c) lim,,,+ x sin (l/x) = 0. [Note: (sin l/x( s 1 for all x > 0, so that
lx sin 11x1 5 1x1 for all positive real numbers x (recall Exercise 5, Article 4.3).]- Use an epsilon-delta argument to prove:
(a) If f(x) = K is a constant function defined on ZR, then lirn,,, f(x) = k, for any
a E IR.
(b) If lirn,,, f (x) = L and K is any real number, then lirn,,, (kf (x)) = kL.- Use an argument similar to the proof in Example 12 and the definitions in
Exercises 4 and 8 of Article 4.3, to prove:
(a) Iflim,,,+ f(x)=L; andlim,,,, g(x)= L,, thenlim,,,+ (f +g)(x)= L, + L,.
(b) If lirn,, , f (x) = L , and lirn,, , g(x) = L,, then lirn,, , (f + g)(x) = L , + L,.- (a) Prove that if lirn,,, f(x) = L where L > 0, then there exists 6 > 0 such that
f (x) > 0 for all x E N'(a; 6). [Recall from Exercise 9 that N1(a; 6) = N(a; 6) - (a}.]
(b) Conclude from (a) that iff is continuous at a and f(a) > 0, then there exists
6 > 0 such that f(x) > 0 for all x E (a - 6, a + 6).
(c) Let f be continuous on an open interval a < x < b containing the real number
x, and suppose f'(x,) > 0. Prove that there exists a neighborhood N(x,; 6) of x,
such that if x E N(x,; 6) and x < x,, then f (x) < f (x,), while if x E N(x,; 6) and
x > x,, then f(x) > f (x,). [Hint: Apply the epsilon-delta definition of limit with
E = f '(xo)/2, using the fact that f '(x,) = lim,,,, (f (x) - f(x,))/(x - x,).] - (a) (Sandwich or Pinching Theorem) Suppose that three functions f, g, and h
are all defined in some neighborhood N(a; r) of a point a, and that f(x) I g(x) I
h(x) for all values of x in this neighborhood, with the possible exception of a itself.
Suppose furthermore that lirn,,, f (x) = lirn,,, h(x) = L. Prove, by using an
epsilon-delta argument, that lirn,,, g(x) = L.
(b) Use (a) to prove that lirn,,, x sin (llx) = 0.
(c) Suppose f is a bounded function (i.e., there exists M > 0 such that I f(x)l M
for all x E R) and g is a function satisfying lirn,,, g(x) = 0. Prove that
lim,,, f (x)g(x) = 0. - Prove that if lirn,,, f (x) = L , and lim,,, g(x) = L,, then lirn,,, (fg)(x) =
L1L2. (Hint: Note that I(fg)(x) - LiL2I S I&)(lf (4 - L1I + IL1 IIdx) - L21,
and that by using the fact that lirn,,, g(x) = L, with E = 1, we can conclude that
Ig(x)I < IL,~ + 1 for all x within some deleted 6 neighborhood of a.)