
“real: chapter_06” — 2011/5/22 — 23:14 — page 15 — #15
Differentiation 6-15
Since x > 0 is arbitrary we have f (x) = cos x − 1 +
x
2
2
> 0or
that cos x > 1 −
x
2
2
.
Theorem 6.3.15 If f is differentiable on [a, b], then all the disconti-
nuities of f
are of second kind.
Proof We shall show that if f
(y+) exists for some y ∈[a, b], then
f
(y+) = f
(y) and similarly if f
(y−) exists for some y ∈[a, b], then
f
(y−) = f
(y) (this will show that there are no discontinuities of first
kind for f
in [a, b]). By definition
f
(y) = lim
h→0
f (y + h) − f (y)
h
.
We are free to allow h → 0 with the condition that h > 0. Applying
the Mean Value Theorem we have
f
(y) = lim
h→0,h>