1. Problem 1 Show that
lim
x→0
sin(x)
=1
x
using an − δ proof.
Solution: One can see that the following inequalities are true for values close to zero, both positive and negative.
| sin(x)| < |x|
|x| < | tan(x)|
This in turn implies that
| cos(x)| < |
sin(x)
|<1
x
On the interval (−π/2, π/2), this implies
cos(x) <
sin(x)
<1
x
Subtracting 1 from both sides, we have
cos(x) − 1 <
sin(x)
−1<0
x
Taking absolute values, again, we have
|
sin(x)
− 1| < |1 − cos(x)|
x
This step is important, since we can show that |1 − cos(x)| goes to zero as x does, that is, the right hand side
can be found in terms of δ.
Since
|1 − cos(x)| = |
1 − cos2 (x)
sin2 (x)
|=
≤ x2
1 + cos(x)
1 + cos(x)
Putting this together we have
|
sin(x)
− 1| < x2
x
Therefore, if |x − 0| < δ then
|f (x) − L| = |
Summing up, if |x| ≤ δ =
√
sin(x)
− 1| ≤ |x|2 ≤ δ 2
x
then
|f (x) − L| = |
sin(x)
− 1| ≤ |x|2 ≤ δ 2 = x
2. Problem 2 Using the results of the previous problem, show that
lim
x→0
sin(sin(x))
x
exists.
Solution: The easiest way is to write the problem as
lim
x→0
sin(sin(x))
sin(sin(x))
sin(x)
= lim
lim
x→0
x→0
x
sin(x)
x
Let u = sin(x), then we have
lim
x→0
sin(sin(x))
sin(u)
sin(x)
= lim
lim
= (1)(1) = 1
u→0
x
u x→0 x
3. Problem 3 Show that
lim sin(1/x)
x→0
does not exists, using an − δ proof.
Solution: The easiest way is a proof by contradiction.
Suppose the limit did exist, then there would be an L such that given an > 0, then |x| < δ would imply
| sin(1/x) − L| < .
Choose an > 0. Find the δ, depending on . We can find an x-value, e.g. x1 = 1/(N π) such that |x1 | < δ, so
that
| sin(1/x1 ) − L| = | sin(N π) − L| < |0 − L| < Similarly, we can find an x-value, e.g. x2 = 1/((2N + 1/2)π) so that |x2 | < δ, so that
| sin(1/x2 ) − L| = | sin((2N + 1/2)π) − L| = |1 − L| < This is a problem, since if < 1/2, L can’t be close to both 0 and 1!
Intuitively, sin(1/x) oscillates to rapidly near x = 0. It takes on values near -1, 0, +1, arbitrarily close to x = 0
so it cannot approach a limit...
4. Problem 4 Show that
√
√
lim [ x + 1 − x] = 0
x→∞
Note: To show that
lim f (x) = L
x→∞
we must show that given any > 0, we can find an N , depending on , such that
|x| > N =⇒ |f (x) − L| < The first step is to multiply by the conjugate
√
√
√
√
√
√
x+1+ x
lim [ x + 1 − x] = lim [ x + 1 − x][ √
√ ]
x→∞
x→∞
x+1+ x
= lim √
x+1−x
√
x+1+ x
= lim √
1
√
x+1+ x
x→∞
x→∞
The critical observation is that this can be estimated in terms of N .
√
So if |x| > N > 4/2 , then
|√
1
1
1
√ < 2√ < 2√
x
x+1+ x
N
1
2
2
√ | < |√ | < √ | < x
x+1+ x
N
This gives the relationship between N and explicitly.
5. Problem 5 The Fibonacci numbers are defined by the relationship
an+1 ≡ an + an−1 , n = 1...∞
with a0 = 1, a1 = 1. What we want to show is that
√
√
√
√
5+ 5 1+ 5 n 5− 5 1− 5 n
(
) +
(
)
a(n) = an =
10
2
10
2
for all integer values of n. This is an indication that we must use induction.
First we show that it is true for n = 1.
√
√
√
√
√
√
5+ 5 1+ 5 0 5− 5 1− 5 0
5+ 5 5− 5
a(1) =
(
) +
(
) =
+
=1
10
2
10
2
10
10
which is true. Now asssume that is true for n ≤ k, that is
√
√
√
√
5+ 5 1+ 5 k 5− 5 1− 5 k
(
) +
(
)
a(k) =
10
2
10
2
√
√
√
√
5 + 5 1 + 5 k−1 5 − 5 1 − 5 k−1
(
)
+
(
)
a(k − 1) =
10
2
10
2
Then we examine
a(k + 1) = a(k) + a(k − 1) =
√
√
√
√
5+ 5 1+ 5 k 5− 5 1− 5 k
(
) +
(
)
10
2
10
2
√
√
√
√
5 + 5 1 + 5 k−1 5 − 5 1 − 5 k−1
+
(
)
+
(
)
10
2
10
2
and by combining common terms,
√
√
√
√
√
√
5 + 5 1 + 5 k−1
1+ 5
5 − 5 1 − 5 k−1
1− 5
=
(
)
(1 +
)+
(
)
(1 +
)
10
2
2
10
2
2
The crucial observation is that
√
√
√
1+ 5 2
1+2 5+5
1+ 5
(
) =
=1+(
)
2
4
2
so
(
Therefore,
√
√
√
1 + 5 k+1
1 + 5 k−1
1+ 5 k
)
=(
)
+(
)
2
2
2
√
√
√
√
5 + 5 1 + 5 k+1 5 − 5 1 − 5 k+1
a(k + 1) =
(
)
+
(
)
10
2
10
2
To get the asymptotics, note that
an
√
√
5+ 5 1+ 5 n
(
10
2 )
=1+
and since
√
√
5− 5 1− 5 n
√ (
√ )
5+ 5 1+ 5
√
1− 5
√ |<1
1+ 5
√
1− 5 n
√ | →0
|
1+ 5
|
hence
√
an
√
5+ 5 1+ 5 n
10 ( 2 )
Therefore c =
√
5+ 5
10
and r =
√
1+ 5
2 .
→1
6. Problem 6 Show that one can compute π by means of the infinite series
1 1 1 1
1
+ − + −
+ ...)
3 5 7 9 11
π = 4 ∗ (1 −
=
∞
X
(−1)n
n=0
1
2n + 1
Solution: The function tan−1 (x) can be written as
Z
Z
1
−1
tan (x) =
dx = (1 − x2 + x4 − x6 + ...)
1 + x2
=x−
x3
x5
x7
+
−
+ ...
3
5
7
Substituting x = 1, we have
π/4 = tan−1 1 = 1 +
1 1 1 1
1
+ − + −
+ ...
3 5 7 9 11
7. Problem 7 Show that the non-alternating series
∞
X
1
2n
+1
n=0
diverges. One can do this several ways. The direct approach is to group terms. If we take k successive terms,
staring with 2N1+1 we have (taking k of the smallest terms as a lower bound)
1
1
1
1
+
+
+
2N + 1 2N + 3 2N + 5 2N + 2k − 1
k
2N + 2k − 1
This will be greater than a fixed constant, e.g. 1/4, if
≥
1
k
≥
2N + 2k − 1
4
that is, if k ≥ N − 1/2. So, letting k = N we have the inequality
1
1
1
1
+
+
+
=
2N + 1 2N + 3 2N + 5 2N + 2k − 1
1
1
1
1
+
+
+
2N + 1 2N + 3 2N + 5 2N + 2N − 1
1
N
≥
≥
4N − 1
4
Since we have an infinite number of these groups, all of which are greater than 1/4, the series diverges.
Another way of demonstrating this is to compare the series to an integral. Using the left hand rule for Riemanns
sums
Left Hand Endpoint
Riemann Sum
y
1
f(x)=1/x
1/2
1/3
1/4
1
This implies that
1+
2
3
4
5
x
∞
∞
X
X
1 1
1
1
1 1
+ + ... =
≥
= + + ...
3 5
2n
+
1
2n
2 4
n=0
n=1
=
1
1 1
1
1 1
1
[1 + + + ...] ≥ [1 + + + ... + ]
2
2 3
2
2 3
N
Z
1 N 1
1
≥
dx = ln(N )
2 1 x
2
As N → ∞, the integral (and therefore the sum) become infinite. This is known as an integral comparison
test.
8. Problem 8 See the file probem8.pdf
9. Problem 9 Define the Heaviside function by
(
0
H(t) ≡
1
if t < 0
if t ≥ 0
Show, uising an − δ argument that limit of H(t) as t → 0 does not exist.
Solution This is very similar to Problem 3. Again, one uses a proof by contradiction.
Suppose the limit did exist, then there would be an L such that given an > 0, then |x| < δ would imply
|H(x) − L| < . But, for any δ > 0 we can find two x values such that we must have |H(x1 ) − L| = |0 − L| < and |H(x2 ) − L| = |1 − L| < . This leads to a contradiction if > 1/2.
10. Problem 10 Show that
lim xx = 1
x→0+
Solution Since xx = (eln x )x = ex ln x , we can use the properties of the limits, and the fact that ex is continuous,
to show that
lim+ xx = lim+ ex ln x = elimx→0+ x ln x
x→0
x→0
To calculate the last limit, we use L’Hôpital’s rule.
lim x ln x = lim
x→0+
x→0+
ln x
1/x
= lim
= lim −x = 0
+
1/x x→0 −1/x2
x→0+
Therefore,
lim xx = elimx→0+ x ln x = e0 = 1
x→0+
You can check this out on a calculator for values very close to 0.0