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Math 10850, Honors Calculus 1
Homework 5
Solutions
General and specific notes on the homework
All the notes from homework 1 still apply! Also, please read my emails from September 6, 13
and 27 with comments on the first four homeworks.
Reading for this homework
Chapters 5 of Spivak.
Assignment
NB: I’m taking these from the 4th ed. of Spivak; the pdf easily found online is the 3th ed.,
and there are some slight differences in the problem sets.
1. Spivak Chapter 5, Question 1 parts (i), (ii), (iv).
Solution (part i): 1 is in the domain of the rational function (x2 − 1)/(x + 1), so the
limit is the value of the function at 1, namely 0.
Solution (part ii): The numerator factors as (x − 2)(x2 + 2x + 4). Since 2 is not in
the domain of the function, it is legitimate to cancel the factors of x − 2 above and
below. This leads to
x3 −8
lim =lim(x2+2x+4)=12,
x→2 x−2 x→2
the latter equality since x2 + 2x + 4 is rational, with 2 in its domain, so the limit is the
value at 2.
Solution (part iv): Viewed as a function of x, with y a constant, the domain of this
function is {x : x 6= y}. This means that we can divide through by x − y without
changing the limit (we are essentially multiplying the function by 1, with 1 written as
(1/(x−y))/(1/(x−y)), which is valid as long as x 6= y). This leads to
n n
lim x −y = lim(xn−1+yxn−2+···+yn−2x+yn−1).
x→y x−y x→y
1
This latter is a rational function (in variable x) with y in the domain, so the limit is
the value of the function at y, that is,
yn−1 +yyn−2 +···+yn−2y+yn−1 or nyn−1.
2. Spivak Chapter 5, Question 3 parts (iii), (iv), (v). (Note that δ will depend on ε.)
Solution (part iii): f(x) = 100/x, a = 1. We claim that the limit is 100.
To prove this, suppose that ε > 0 is given. We want to find δ such that whenever
0 < |x −1| < δ, we have |100/x−100| < ε.
Now |100/x−100| < ε is equivalent (after a little algebra) to |x − 1|/|x| ≤ ε/100.
Choose δ < 1/2. Then 0 < |x − 1| < δ implies x ∈ (1/2,3/2), so |x| > 1/2 and
1/|x| < 2.
Choose also δ < ε/50. Then 0 < |x−1| < δ implies |x−1| < ε/50.
To get both conditions to hold, we choose δ < min{1/2,ε/50}, say for concreteness
δ = min{1/4,ε/100}; note that δ > 0.
Forthisδ, or any δ < min{1/2,ε/50}, wehavethatif0 < |x−1| < δ then|100/x−100| <
2(ε/50) = ε.
This proves that limx→1100/x = 100.
Solution (part iv): f(x) = x4, arbitrary a. We claim that the limit is a4.
To prove this, suppose that ε > 0 is given. We want to find δ such that whenever
4 4
0 < |x −a| < δ, we have |x −a | < ε.
4 4 3 2 2 3
Now|x −a |<εisequivalent (after a little algebra) to |x−a||x +ax +a x+a | ≤ ε.
Choose δ < 1. Then 0 < |x − a| < δ implies x ∈ (a − 1,a + 1), so x < a + 1, so
|x| < |a| + 1, so
3 2 2 3 3 2 2 3 3 2 2 3
|x +ax +a x+a | ≤ |x| +|a||x| +|a| |x|+|a| < (|a|+1) +|a|(|a|+1) +|a| (|a|+1)+|a| .
3 2 2 3
Choose also δ < ε/((|a|+1) +|a|(|a|+1) +|a| (|a|+1)+|a| ). Then 0 < |x−a| < δ
3 2 2 3
implies |x − a| < ε/(|a| + 1) + |a|(|a| + 1) + |a| (|a| + 1) + |a| .
3 2
To get both conditions to hold, we choose δ < min{1,ε/(|a| + 1) + |a|(|a| + 1) +
2 3 3 2
|a| (|a| + 1) + |a| }, say for concreteness δ = min{1/2,ε/(2((|a| + 1) + |a|(|a| + 1) +
2 3
|a| (|a| + 1) + |a| ))}; note that δ > 0.
3 2 2 3
For this δ, or any δ < min{1,ε/((|a|+1) +|a|(|a|+1) +|a| (|a|+1)+|a| )}, we have
that if 0 < |x − a| < δ then
4 4 3 2 2 3 3 2 2 3
|x −a | = |x−a||x +ax +a x+a | < |x−a|((|a|+1) +|a|(|a|+1) +|a| (|a|+1)+|a| ) < ε.
4 4
This proves that lim x =a .
x→a
Solution (part v): f(x) = x4 +1/x, a = 1. We claim that the limit is 2.
2
To prove this, suppose that ε > 0 is given. We want to find δ such that whenever
0 < |x −1| < δ, we have |x4 +1/x−2| < ε.
Now |x4 +1/x−2| = |(x4 −1)+(1/x−1)| ≤ |x4 −1|+|1/x−1|, so it is enough to
find a δ > 0 such that whenever 0 < |x − 1| < δ, we have both |x4 − 1| < ε/2 and
|1/x−1| < ε/2.
3 2 2
From part (iv) we know that if δ < min{1,(ε/2)/((|1| +1) +|1|(|1|+1) +|1| (|1|+
3 4
1) +|1| )} = min{1,ε/30} then 0 < |x−a| < δ implies |x −1| < ε/2.
From part (iii) we know that for δ < min{1/2,ε/50}, we have 0 < |x−1| < δ implies
|100/x−100| < ε so |1/x−1| < ε/100, from which it follows that if δ < min{1/2,ε},
we have 0 < |x−1| < δ implies |1/x−1| < ε/2.
Sotoshowlim (x4+1/x) = 2, givenε > 0weshouldtakeδ < min{1,ε/30,1/2,ε/50} =
x→1
min{1/2,ε/50}.
3. Spivak Chapter 5, Question 7.
Solution: The graph below gives an example of such a function: (for the record, it is
a portion of the graph of f(x) = p|x|).
Take ℓ = a = 0, and ε = δ = 1. It is certainly the case that
0 < |x| < 1 =⇒ |f(x)| < 1.
But it is not the case that
0 < |x| < 1/2 =⇒ |f(x)| < 1/2;
the best that we can ensure from the imposition 0 < |x| < 1/2 is |f(x)| < .7; to force
|f(x)| < 1/2 we need to demand that 0 < |x| < 1/4.
3
4. Spivak Chapter 5, Question 8.
Solution (part a): Yes. Consider, for example,
f(x) = 1 if x > 0
−1 if x < 0
and
g(x) = −1 if x > 0
1 if x < 0.
Certainly lim f(x) and lim g(x) do not exist. But (f + g)(x) = 0 unless x = 0
x→0 x→0
(at which point the sum is undefined), so limx→0(f +g)(x) = 0.
Solution (part b): Yes. If lim f(x) exists, and lim (f(x)+g(x)) exists, then by
x→a x→a
the sum-product-reciprocal theorem,
lim((f(x)+g(x))−f(x)) = limg(x)
x→a x→a
exists.
Solution (part c): No. If lim (f(x) + g(x)) existed then (by part b) lim g(x)
x→a x→a
would also exist, a contradiction.
Solution(partd): Itistemptingtosay“yes”. Iflim f(x)exists, and lim f(x)g(x)
x→a x→a
exists, then by the sum-product-reciprocal theorem,
lim(f(x)g(x))/f(x)) = limg(x)
x→a x→a
should exist; but this assumes that lim f(x) is not zero. So to find a counter-example,
x→a
we need to find functions f and g, and an a, with lim f(x) = 0, lim f(x)g(x)
existing, and lim g(x) not existing. x→a x→a
x→a
Taking f to be the constant 0 function, g to be the function g(x) = sin(1/x) and a = 0
works nicely.
5. Spivak Chapter 5, Question 9.
Solution: An implicit assumption here (contained in the writing down of the two
expressions) is that lim f(x), lim f(a+h)bothexist. Supposethatlim f(x) =
x→a h→0 x→a
L. We will argue that lim f(a+h)=L.
h→0
Given ε > 0, we know that there is δ > 0 such that 0 < |x−a| < δ implies |f(x)−L| < ε
(this because lim f(x) = L). We claim that the same δ works (for this ε) to show
x→a
lim f(a+h)=L. Indeed, we need to show that when
h→0
0 < |h| < δ
then
|f(a +h)−L| < ε.
4
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