Calculus
Instructor: Mohamed Omar
Assignment 2
Due: Monday Oct. 15, 2012. 11:59pm
Math 1a Section 1
1. Let n ≥ 1 be a positive integer. Prove
13 + 23 + · · · + n3 =
n2 (n + 1)2
.
4
2 2
Solution: When n = 1, the left side is 1 and the right side is 1 42 = 1. Now assume
that the equality holds from some positive integer n. We claim the statement holds
for n + 1. Indeed,
n2 (n + 1)2
+ (n + 1)3
4 n2
+n+1
= (n + 1)2
4
2
2 n + 4n + 4
= (n + 1)
4
2
(n + 1) (n + 2)2
=
.
4
13 + 23 · · · + n3 + (n + 1)3 =
Thus by Mathematical Induction, the statement holds for all positive integers n.
2. Prove that for any integer n ≥ 1,
(cos(θ) + i sin(θ))n = cos(nθ) + i sin(nθ),
where i2 = −1.
Solution: Equality clearly holds for n = 1. Now suppose n ≥ 2 is a positive integer,
and (cos(θ) + i sin(θ))n = cos(nθ) + i sin(nθ). Then
(cos(θ) + i sin(θ))n+1 = (cos(θ) + i sin(θ))n (cos(θ) + i sin(θ))
= (cos(nθ) + i sin(nθ)) · (cos(θ) + i sin(θ))
= (cos(nθ) cos(θ) − sin(nθ) sin(θ)) + i(sin(nθ) cos(θ) + sin(θ) cos(nθ))
= cos((n + 1)θ) + i sin((n + 1)θ)
Thus by Mathematical Induction, the statement holds for all positive integers n.
3. For each of the following, prove or disprove the limit exists.
(a)
lim (2x − 1).
x→3
1
Solution: Let > 0. Choose δ = 2 . Then if x satisfies 0 < |x − 3| < δ, then
|(2x − 1) − 5| = |2x − 6| = 2|x − 3| < 2δ = .
Thus, by the definition of a limit,
lim (2x − 1) = 5.
x→3
(b)
lim x2 .
x→3
Solution: We claim limx→3 x2 = 9. Indeed, given > 0, we choose δ > 0 to be
n o
,1
δ = min
7
Then if 0 < |x − 3| < δ, we have
|x2 − 9| = |x − 3||x + 3| = |x − 3|(|x − 3 + 6|) ≤ |x − 3|(|x − 3| + 6) < 7|x − 3| < .
The last inequality comes from the fact that |x − 3| < /7. The second last
inequality comes from the fact that |x − 3| < 1.
(c)
lim
x→0
1
.
x
Solution: Suppose otherwise. That is, suppose limx→0 x1 = L for some real
number L. We’ll assume L ≥ 0; the argument for when L < 0 is similar. Let
= 1. Then for any arbitrary δ > 0 it suffices to find a value of x with |x| < δ
such that f (x) > L + 1. choose
δ
1
x = min
,
.
2 2(L + 1)
Certainly x satisfies |x| < δ, and we see that
f (x) =
1
≥
x
1
1
2(L+1)
= 2(L + 1) > L + 1.
We conclude
lim
x→0
1
x
does not exist.
4. Give examples to show that the following definitions of limx→a f (x) = ` are not
correct.
2
(a) For all δ > 0 there is an > 0 such that if 0 < |x − a| < δ then |f (x) − `| < .
Solution: This statement is true for any bounded function, regardless if it’s
continuous or not. As an example, consider the function f (x) defined by f (x) = 1
if x ≥ 0 and f (x) = −1 if x < 0. For this function, limx→0 f (x) does not exist.
However, given any δ > 0, if we pick = 3 we have that if x is in the range
of values for which 0 < |x − 0| < δ, then |f (x) − f (1)| is at most 2, and hence
|f (x) − f (1)| < 3, so this definition would imply limx→0 f (x) = 1, which is false.
(b) For all > 0 there is a δ > 0 such that if |f (x) − `| < then 0 < |x − a| < δ.
Solution: We will show that there are functions that have limits at a specific
point, but violate the given definition. Consider the function f (x) = 2 for every
x. Certainly, limx→1 f (x) = 2. However, the given definition implies does not
hold. Indeed, choose > 0, and let δ > 0 be any thing. Then, in fact, every
single value of x satisfies |f (x) − 2| < , but certainly if we make x very large it
will not satisfy |x − 0| < δ.
5. Define the function f (x) on the interval [0, 1] by
0
if x is irrational
f (x) =
1/n
if x ∈ Q and x = m/n in reduced form
Determine, with proof, whether or not
lim f (x) = f
9
x→ 2012
9
2012
.
Also, determine, with proof, whether or not
lim f (x) = f
x→ √1
2
1
√
2
.
Solution: We first consider
lim f (x) = f
9
x→ 2012
9
2012
.
We claim this statement is not true. Observe that there are only finitely many rational
numbers in the interval [0, 1] whose denominators are greater than 2012. Let M be
9
the minimum distance from 2012
to any of these rational numbers. Then for any x in
9
M
9
1
the interval 2012 − 2 ≤ x ≤ 2012
+M
2 we have f (x) ≤ 2013 .
1
1
Now let = 2012
− 2013
. Now given any δ > 0, choose a real number x such that that
9
9
M
9
9
is in the interval 2012 − M
2 ≤ x ≤ 2012 + 2 and the interval 2012 − δ ≤ x ≤ 2012 + δ.
1
Then 0 ≤ f (x) ≤ 2013
, so
9
f (x) − f
= f (x) − 1 = 1 − f (x) ≥ 1 − 1 = .
2012
2012 2012
2012 2013
3
Now consider
lim f (x) = f
x→ √1
2
1
√
2
= 0.
We prove this statement is in fact true. Choose > 0. Let N be any integer such
that N1 < . There are only finitely many fractions whose denominators are less than
or equal to N . Let M be the minimum distance from any of these fractions to √12 .
Let δ = M . Then for any x satisfying 0 < |x − √12 | < M , we have
f (x) − f √1 = |f (x) − 0| = f (x) < 1 < .
N
2 6. Compute the following limits (you don’t need to use the definition of a limit at all):
(a)
√
√
lim ( x + 1 − x).
x→∞
Solution: Observe that
√
√
(x + 1) − x
1
x+1− x= √
√ =√
√ .
x+1+ x
x+1+ x
Thus
√
√
lim ( x + 1 − x) = lim √
x→∞
x→∞
1
√ = 0.
x+1+ x
1
1
√ ≤ √ , and both limx→0 0 = limx→0
2
x
x+1+ x
conclude by the Squeeze Theorem
√
√
lim ( x + 1 − x) = 0.
Now since 0 ≤ √
1
√
2 x
= 0, we
x→∞
(b)
lim
x→0
tan(3x)
.
x
Solution: The argument in the limit is
3 · lim cos(3x) · lim
x→0
x→0
sin(3x)
tan(3x)
=
. Thus, our limit is
x
x cos(3x)
sin(3x)
= 3 · 1 · 1 = 3.
3x
(c)
1
lim (7x + 2x ) x .
x→∞
Solution:
Notice that 7x ≤ 7x + 2x ≤ 2(7x ), the last inequality being true because 2x < 7x
for positive x (which is sufficient to look at because we want to know about the
limit as x goes to infinity). Thus
1
1
1
(7x ) x ≤ (7x + 2x ) x ≤ [2(7x )] x .
4
Simplifiying, we have
1
1
7 ≤ (7x + 2x ) x ≤ 7 · 2 x .
1
Since limx→0 2 x = 1 (because limx→0
1
2 x = 7. Thus
1
x
= 0), we have that limx→0 7 = limx→0 7 ·
1
lim (7x + 2x ) x = 7.
x→∞
5