MATH 41
TA Section Notes for Thu 04 Oct 07
Admin:
• notes on my homepage math.stanford.edu/∼jasonlo
Today:
• ε-δ definition of limits
• squeeze theorem
• continuity
• intermediate value theorem
1
ε-δ definition of Limits
Appendix D, Ex 1 Rewrite the condition
|
1
− 0.5 |< 0.2
x
and then expalain what happens when δ is large or small.
Appendix D, Ex 9 Recall the definition of limit on p.A31: we write
lim f (x) = L
x→a
if for every number ε > 0 we can find a corresponding number δ > 0 such that
|f (x) − L| < ε
whenever
0 < |x − a| < δ
According to this definition, if we want to prove
lim (3x − 2) = 4,
x→2
then for every number ε > 0, we need to produce a corresponding number δ > 0
such that
|(3x − 2) − 4| < ε whenever 0 < |x − 2| < δ.
Now we can rewrite the first inequality above:
|(3x − 2) − 4| < ε ⇔ |3x − 6| < ε
⇔ |3(x − 2)| < ε
⇔ 3|x − 2| < ε
ε
⇔ |x − 2| <
3
1
So we see that if we put δ := 3ε , we should be fine. But let’s check it (you must
do this checking in an exam!):
whenver 0 < |x − 2| < δ = 3ε , we get
|x − 2| <
ε
⇒ 3|x − 2| < ε
3
⇒ |3(x − 2)| < ε
⇒ |3x − 6| < ε
⇒ |(3x − 2) − 4| < ε
So for every ε > 0, if we choose δ := 3ε then we’ll be fine. Hence the limit is as
claimed.
Note. You can look at Example 3, p.A32 for another example.
2
Squeeze Theorem
Section 2.3, Ex 27 Just an application of the squeeze theorem.
3
Continuity
Section 2.4, Ex 3 (a) The function is not defined at x = −4, so there’s no
need to talk about whether it is continuous or distontinuous there.
Discontinuities at: x = −2, 2, 4. Recall that a function f (x) is continuous
at x = a if limx→a f (x) = f (a). Implicitly (cf. p.117), this requires checking 3
things:
• f (x) is defined at x = a
• limx→a f (x) exists (i.e. the limit from either side exists, and these two
limits are equal)
• limx→a f (x) equals f (a)
Section 2.4, Ex 31 (Perhaps there’s no time to go through this in section.)
We know that x + 2, ex and 2 − x are all continuous functions on all of R.
So discontinuities of f (x) could only occur at where we ”glue” pieces of these
functions together: x = 0, 1.
From the definition of f , limx→0− f (x) = 0 + 2 = 2, while f (0) = e0 = 1 and
limx→0+ f (x) = e0 = 1. Hence f is discontinuous at x = 0, but is continuous
from the right there.
From definition of f , limx→1+ f (x) = 2 − 1 = 1, while limx→1− f (x) = e1 =
e ≈ 2.7183, and f (1) = e1 = e. So f (x) is discontinuous at x = 1, but is
continuous from the left there.
Sketch the graph. You could’ve sketched the graph first if it helps you work
out the above answers.
2
Section 2.4, Ex 33 The potential discontinuity is at x = 2. As stated above,
there are three things we need to check in order for f to be continuous at x = 2:
• f (x) is defined at x = 2.
• limx→2+ f (x) certainly exists since x3 − cx is a polynomial, and it equals
23 − c · 2 = 8 − 2c.
limx→2− f (x) = c · 22 + 2 · 2 = 4c + 4.
We need the two one-sided limits to agree, i.e. we need
8 − 2c = 4c + 4, ⇔ 6c = 4 ⇔ c = 2/3.
So when c = 2/3, limx→2 f (x) exists and equals 8 − 2c = 8 −
4
3
= 20/3.
• Lastly, we need to check limx→2 f (x) = f (2), which indeed is true.
So when c = 2/3, f is continuous on all of (−∞, ∞).
4
Intermediate Value Theorem
Theorem 4.1 (The Intermediate Value Theorem, p.124). Suppose that
f is continuou on the closed interval [a, b] and let N be any number between
f (a) and f (b), where f (a) 6= f (b). Then there exists a number c in (a, b) such
that f (c) = N .
Draw two diagrams to illustrate this: when f (a) < f (b) and when f (a) >
f (b).
Section 2.4, Ex 39 We want to work out an x such that cos x = x. So if we
write f (x) = cos x − x, then what we are looking for is an x such that f (x) = 0.
Now consider the values of f at x = 0, π: f (0) = cos 0 − 0 = 1, whereas
f (π) = cos π − π 3 = −1 − π 3 < 0. Since f is continuous on [0, π], and 0 is a
number between f (π) and f (0) (these two are clearly different numbers), the
intermediate value theorem gives us immediately that there is an x in (0, π)
such that f (x) = 0, so we’re done.
Section 2.4, Ex 47 A number x with such a property would satisfy the
equation x = 1 + x3 . So consider f (x) = 1 + x3 − x = x3 − x + 1, which is a
cubic equation. Note what happens when x → ∞, and when x → −∞. For
example, we can consider x = 10, −10.
As above, we compute f (10) = 1000 − 10 + 1 = 991, whereas f (−10) =
−1000 + 10 + 1 = −989. Now we know f is continuous on [−10, 10], and 0 is
a number between f (−10) and f (10) (and clearly f (−10) 6= f (10)). Hence by
the intermediate value theorem, we know there is a number x in (−10, 10) such
that f (x) = 0; this x would satisfy the properties we want.
3