MthT 430 Projects Chapter 6a Solution

Limits

1.

Let f(x) be a function such that

•

domain( f) = [0,1).

•

For all x (in [0,1)), 0 ≤ f(x) < 1.

•

The function f is increasing on [0,1).

Show that there is a number L, 0 ≤ L ≤ 1, such that

lim x → 1−  f(x) = L.

Hint: Construct a binary expansion for L.

2.

Discuss the continuity of the function described on p. 97 and whose graph is sketched in FIGURE 14.

3.

Prove: If g is continuous at a, g(a) ≠ 0, then there is a δ > 0 for which (a − δ,a + δ) is contained in the domain of [1/g].

Solution. For every ϵ > 0, there is some δ > 0 such that, for all x, if |x −a| < δ, then |g(x) − g(a)| < ϵ.

Let ϵ = |g(a)|. Then there is a δ > 0 such that for |x−a| < δ, |g(x) − g(a)| < |g(a)|. Thus for a −δ < x < a +δ, g(a) − |g(a)| < g(x) < g(a) + |g(a)|; if g(a) > 0, 0 < g(x) < 2 g(a); if g(a) < 0, 2 g(a) < g(x) < 0. In either case, for a − δ < x < a +δ, g(x) ≠ 0, and x is in the domain of 1/g.

Another Solution. For every ϵ > 0, there is some δ > 0 such that, for all x, if |x −a| < δ, then |g(x) − g(a)| < ϵ.

Let ϵ = |g(a)|. Then there is a δ > 0 such that for |x−a| < δ, |g(x) − g(a)| < |g(a)|. Thus for a −δ < x < a +δ, |g(x)| = |g(a) + (g(x) − g(a)) | ≥ |g(a| − |g(x) − g(a)| > 0. Here we have used the triangle inequality in the form |A ±B| ≥ |A| − |B|.

Thus, for a − δ < x < a +δ, g(x) ≠ 0, and x is in the domain of 1/g.

Good Variation ... ϵ = |g(a)| ... If g(a) > 0, ... for a − δ < x < a + δ, g(x) ∈ (g(a) − ϵ, g(a) + ϵ) = (0, 2 g(a)) and g(x) ≠ 0. ...

---

File translated from T_EX by T_TH, version 4.05.
On 18 Sep 2014, 21:18.