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. ...
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