MAT 135 (Section L5102) - Lecture 4
Announcements
1. The MAT135 Test will be held on Monday, October 24
(7:45 to 9:15 pm). Any student with a time conflict that
night should e-mail Anthony Lam immediately
(lam@math.utoronto.ca), for special arrangements.
2. Some old tests of MAT135Y are posted on the MAT135
Website (http://www.math.toronto.edu/lam/).
3. (Belated) Shana Tova!
4. (Not So Belated) Happy Durga Puja!
5. Who did not vote?
Review of Lecture 3: Moral for finding limits
Now that we know what continuous functions are (we covered it in
the class today), the general procedure to find limx→a f (x) is as
follows:
1. Check if a ∈ dom(f ). If a ∈ dom(f ) and f is continuous at a,
then limx→a f (x) = f (a), i.e. you can just plug in x = a !
Therefore if f is a nice function, e.g. polynomial or rational or
trigonometric or exponential or logarithmic function and
a ∈ dom(f ), then to find limx→a f (x), you only have to see if
a is in the domain of f , and if a is in the domain, plug in a for
x in the expression for f .
2. If a 6∈ dom(f ) or you are not sure if f is continuous at a, then
you try to find limx→a− f (x) and limx→a+ f (x). The limit
exists only if both these one-handed limits exist and are equal
to each other.
Review of Lecture 3 Contd.
f is continuous at a iff limx→a f (x) = f (a).
Remember: f is continouous at a only if ALL of the following 3
conditions hold:
1. a ∈ dom(f )
2. limx→a f (x) exists, and
3. limx→a f (x) = f (a).
Squeeze Theorem:
If f (x) ≤ g (x) ≤ h(x) for all x near a and
limx→a f (x) = limx→a h(x) = L, then limx→a g (x) also equals L.
Intermediate Value Theorem:
If f is continuous on [a, b] and f (a) 6= f (b) and L is between f (a)
and f (b) (but not equal to any of f (a) or f (b)), then there exists
c ∈ (a, b) such that f (c) = L.