Homework 6 Key
• Let f be a continuous function on R. Suppose f ′ (x) exists for all x ̸= 0
and f ′ (x) → 7 as x → 0. Does it follow that f ′ (0) exists?
Consider the expression
f (x) − f (0)
.
x−0
By the MVT we have f (x) − f (0) = f ′ (z) for some z between 0 and x.
So
f (x) − f (0)
= f ′ (z).
x−0
If we limit x → 0 we see also that z → 0, and so f ′ (z) → 7. But
the limit as x → 0 on the left above is just f ′ (0), so we conclude that
f ′ (0) = 7.
Or you can use L’Hopital’s rule: Consider the limit
lim+
x→0
f (x) − f (0)
.
x−0
This is an indeterminate form since limx→0+ f (x) = f (0) (f is continuous.) L’Hopital’s rule applies and so let’s look at
lim+
x→0
We conclude that
lim+
x→0
f ′ (x)
= 7.
1
f (x) − f (0)
= 7.
x−0
Similar analysis shows that
lim−
x→0
f (x) − f (0)
= 7.
x−0
It follows that f ′ (0) = 7.
• Define a function f : R → R as
{ 2
x sin(1/x), x ̸= 0
f (x) =
0,
x=0
1
If x ̸= 0 then the usual rules for derivatives apply and we find that
f ′ (x) = 2x sin(1/x) − cos(1/x). At x = 0 we have
f (y) − f (0)
= lim y sin(1/y) = 0
y→0
y→0
y−0
f ′ (0) = lim
since −1 ≤ sin ≤ 1. Thus
{
2x sin(1/x) − cos(1/x), x ̸= 0
′
f (x) =
0,
x=0
But it’s easy to see that limx→0 f ′ (x) does not exist, hence f ′ is discontinuous there.
• Define a function f : R → R as
{
x + 2x2 sin(1/x), x ̸= 0
f (x) =
0,
x=0
If x ̸= 0 then the usual rules for derivatives apply and we find that
f ′ (x) = 1 + 4x sin(1/x) − 2 cos(1/x). At x = 0 we have
f (y) − f (0)
= lim 1 + 2y sin(1/y) = 1
y→0
y→0
y−0
f ′ (0) = lim
since −1 ≤ sin ≤ 1. Thus
{
1 + 4x sin(1/x) − 2 cos(1/x), x ̸= 0
′
f (x) =
1,
x=0
Now we show that even though f ′ (0) > 0, f is not monotone increasing
in any neighborhood of x = 0. We can find points x0 arbitrarily close
1
to x = 0 such that f ′ (x0 ) > 0. For example, let x0 = 2kπ
where k ∈ Z.
Then
f ′ (x0 ) = −1.
Since f ′ (x) is continuous away from x = 0, there must be some interval I + of the form I + = (x0 − δ, x0 + δ) such that f ′ (x) < 0 for
x ∈ I + . On this interval f is strictly decreasing, by Theorem 5.11.
Note, to conclude that f is strictly decreasing we need f ′ < 0 ON AN
INTERVAL.
By the way, the same argument works for finding an interval on which
1
f is strictly increasing; use x0 = (2k+1)π
and find f ′ (x0 ) = 3. So f is
neither increasing nor decreasing in any neighborhood of 0.
2