Math 101 Exam Study Questions: Chapters 1-2
1. Limit of a product
Complete the proof of Theorem 2.6 in the case L1 and L2 are both zero.
2. Limit of a composite function revisited
Suppose f and g are both functions from R to R. If g is continuous at a, an
Math 101: Selected solutions, Homework 9
Proposition. Suppose for every
> 0 there exists > 0 such that
x, y (b , b) = |f (x) f (y )| < .
Then there exists L such that
lim f (x) = L.
Lets rst nd L by constructing a Cauchy sequence as follows.
Math 101 Homework 12: Selected solutions
6.1 Problem 9
Let C be the set of continuous functions with domain [0, 1] and dene d : C C R by
|f (x) g (x)|dx.
d(f, g ) =
We will prove d denes a metric on C .
The function d is well-dened because |f (x) g (x
Math 101 Study Questions: Chapters 3-4
1. Continuous maps and open/closed intervals
Determine whether each of the following maps exist by either giving an example or proving such a map is impossible.
(a) A continuous function f : [0, 1] R whose image is e
Math 101 Study Questions: Chapters 5-6
1. Integrals and the L1 metric
Suppose x0 is some point in [0, 1] and y0 is any point in R. Dene the function
h : [0, 1] R by:
y0 if x = x0
Use either the Darboux or Riemann integral to prove that
Math 101 Homework 4: some solutions
Problem for 2.3
Determine whether the function below is continuous at a = 0; if it is not,
determine whether it is continuous from the left, right, or neither.
if x > 0
f (x) = 0
if x = 0
x cos(1/x) if x < 0
Math 101 Homework 1: Partial Solutions
3. Prove that (a)(b) = ab for arbitrary elements a, b in a eld.
This is Theorem 1.7(iii); in class we proved Theorem 1.7(i-ii). By Part (i),
(a)(b) = [(a) b]. By Part (ii), (a) b = (ab). Combining these
Math 101 Homework 3: Partial solutions
2.1 Problem 21
The function f (x) is dened to be 1 if x is rational and 0 if x is irrational.
We prove that f is not continuous at any value of x.
By denition, f is continuous at a if and only if: for any > 0, there
Math 101: Selected solutions, Homeworks 6-8
Hw 6: Problem 1b
Suppose g : R R equals 1 on all rational points and g is continuous. Towards contradiction, suppose g (x) = 1 for some x R. Then we may let = |g (x) 1| > 0 and nd such
that for any x (x , x + ),
Math 101 Homework 2: Partial solution
1.4 Problem 3
Prove n (2i1) = n2 for any natural number n. We proceed by induction.
Because 1=1 (2i 1) = 2 1 = 1 = 12 , the statement is true for n = 1.
Now assume it is true for n = k . Then
(2i 1) =
Math 101: Selected solutions, Homework 10
4.1 Problem 15
Let us prove by induction that fn has exactly n/2 derivatives if n is even, and (n 1)/2
derivatives if n is odd, where
fn (x) =
I nd it easiest to simultaneously prove that gn