Homework 2 Solutions
1. (a) Find the area of a regular hexagon inscribed in a circle of radius 1. Then, find the area of a regular hexagon circumscribed about a circle of radius 1. Use these calculations to estimate the area of a circle of radius 1. Then, use these calculations to give a decimal approximation of π.
When a regular hexagon is inscribed in a circle of radius 1, the hexagon can be divided into 6 triangles as shown below by joining each of the vertices of the hexagon to the center of the circle. Let 4ABO be one such triangle as below. Then angle ∠AOB has measure equal to 2π/6 since each of the angles of the 6 triangles at the point O are congruent. We compute the area of one of these triangles as shown. Let M be the the midpoint of AB. Since 4AM O is a right triangle, we can use trigonometry to find the lengths |AM | and |M O|. Let θ denote the measure of the angle ∠M OA. Then, since |OA| = 1 (being equal to the radius of the circle),
|AM | = sin θ and |M O| = cos θ.
Therefore, the area of triangle 4AOB is equal to (1/2)|AB||M O| = (1/2)(2 ∗ sin θ)(cos θ) = 1
2sin (2θ), the last equality following from the double angle formula
2 sin α cos α = sin (2α).
The area of the hexagon is 6 times the area of the triangle 4AM O. Thus, if we denote the area of the hexagon by A6 and we use the
observation that θ = π/6, then we see that A6 = 6 ·
1
2sin (2π/6) = 3 sin (π/3) = 3√3
This gives us a rough estimate of π. The area of the unit circle is equal to π and the area of the hexagon gives us a lower bound. The computation of the area of a circumscribed hexagon can be computed in a similar manner. Here is a shortcut. It is not hard to compute that the length of one of the sides of a circumscribed hexagon is equal to 2/√3. Therefore the circumscribed hexagon is similar to the inscribed hexagon and the latter is the same as the former scaled by a factor of 2/√3. It follows that the area of the circumscribed hexagon will be (2/√3)2 times the area of the inscribed hexagon. (If lengths are scaled by a factor c > 0, then area is scaled by a factor of c2.) Thus, the area of the circumscribed hexagon is equal to (3/4)(3√3/2) = 2√3.
(b) Find the area Anof a regular n-gon inscribed in a circle of radius
1. What is the value of limn→∞An? Can you justify your answer
analytically (e.g. using the methods of 2.3)?
To solve this problem, proceed exactly as in part (a) above. The only change is that here θ = π/n. Therefore,
An= n ·
1
2sin (2π/n) = n
2sin (2π/n).
The limit of Anas n increases is equal to π since increasing the number
of sides of the polygon will fill in the area of the circle will less and less error.
If n = 96, we see that A96≈ 3.13935. (This is the origin of π ≈ 3.14.)
The reason why n = 96 is a good choice is due to the fact that we have an exact way of computing sin 2π/96. The trick is to use the half angle formulas:
cos2(θ/2) = 1
2(1 + cos θ) and sin
2(θ/2) = 1 2(1 − cos θ). It follows that sin (π/48) = v u u u t1 2(1 − v u u t1 2(1 + s 1 2(1 + √ 3 2 ))). This simplifies to the much more appealing equation
sin (π/48) = 1 2 s 2 − r 2 + q 2 +√3.
2. In this problem, we will study the properties of limits in Theorem 2.2 of section 2.3. Compute each of the following limits and clearly indicate which properties you are using at each step of your solution.
(a) lim x→1 x 2− 3x + 2 • By Theorem 2.1.1, limx→12 = 2. • By Theorem 2.1.2, limx→1x = 1. • By Theorem 2.1.3, limx→1x2 = 12= 1.
• By Theorem 2.2.1 (scalar multiple), limx→1−3x = −3(limx→1x) = −3 · 1 = −3.
• Finally, using two applications of Theorem 2.2.2 (sum), we have that lim x→1 x 2− 3x + 2 = lim x→1x 2+ lim x→1−3x+ limx→12 = 1−3+2 = 0. (b) lim x→−1 x2+ 2x + 1 x2+ 3x + 2
• Factor the numerator an denominator of the expression inside the limit operator above:
(x + 1)(x + 1) (x + 2)(x + 1).
When we compute a limit at x = −1, the values of x approach but never equal -1. Therefore it is valid to cancel the factors of (x + 1) and say that
lim x→−1 x2+ 2x + 1 x2+ 3x + 2 = limx→−1 x + 1 x + 2.
(Here we are using Theorem 2.7 (functions which agree at all but one point).)
• By Theorem 2.1.1, limx→−11 = 1 and limx→−12 = 2.
• By Theorem 2.1.2, limx→−1x = −1.
• By Theorem 2.2.2 (sum), we have that lim
x→−1(x + 1) = limx→−1x + limx→−11 = −1 + 1 = 0 and
lim
• Finally, by Theorem 2.2.4 (quotient), we have that lim x→−1 x + 1 x + 2 = limx→−1(x + 1) limx→−1(x + 2) = 0 1 = 0. (c) lim x→0 1 − cos x sin x
• If x is close to zero (more precisely, if |x| < π/2), then (1 + cos x) 6= 0. Therefore, 1 − cos x sin x = 1 − cos x sin x · 1 + cos x 1 + cos x. • The above simplifies to
1 − cos2x sin x(1 + cos x) =
sin2x sin x(1 + cos x).
• When we compute the limit as x approaches zero, the value of x is never equal to zero. Therefore, it is valid to cancel the factors of sin x in the above and say (by Theorem 2.7) that
lim x→0 1 − cos x sin x = limx→0 sin x 1 + cos x.
• By Theorems 2.6.1 and 2.6.2, we have that limx→0sin x = sin 0 = 0 and limx→0cos x = cos 0 = 1.
• By Theorem 2.1.1, we have that limx→01 = 1.
• By Theorem 2.2.2 (sum), we have that lim
x→0(1 + cos x) = limx→01 + limx→0cos x = 1 + 1 = 2
• Finally, by Theorem 2.2.4 (quotien), we have that lim x→0 sin x 1 + cos x = limx→0sin x limx→0(1 + cos x) = 0 2 = 0.
The approach taken in these exercises in problem #2 is what might be called an “axiomatic approach”. The point of this exercise was to try to understand how often we implicitly use properties and theorems about limits when we make computations. The upshot is that one these properties and theorems are established, computation becomes as easy as doing arithmetic in many cases.
3. This is an example of a challenging problem. Only a few exam or quiz problems will be “challenging”, but here is a chance to see if you’re ready! Solve exercise 70 in section 2.2. Your solution should begin with a clear statement the problem. Explain how to simplify
|x + 1| − |x − 1| x
algebraically if −1 ≤ x < 0 or 0 < x ≤ 1.
To simplify the expression algebraically, we use the fact that |x| = x if x ≥ 0 and |x| = −x if x ≤ 0. From this it follows that |x + 1| = x + 1 if x ≥ −1 and |x + 1| = −(x + 1) if x ≤ −1. Similarly, |x − 1| = x − 1 if x ≥ 1 and |x − 1| = −(x − 1) if x ≤ 1.
If we compute a limit as x approaches 0, then we only need to consider x-values close, but not equal to 0. Therefore, we may restrict our attention to x-values such that −1 ≤ x ≤ 1 and x 6= 0. Using this, we have that |x + 1| − |x − 1| x = (x + 1) − −(x − 1) x = 2x x = 2
provided that −1 ≤ x ≤ 1 and x 6= 0. Therefore, the above limit is equal to 2. This can also be observed by testing values close to 0. Be careful to test both values to the right and the left when computing limits of expressions which involve the absolute value. You can also find this limit by sketching the graph, although this is more difficult for this problem. The graph is quite interesting however. Notice that if x ≥ 1, then |x + 1| − |x − 1| x = (x + 1) − (x − 1) x = 2 x. Similarly, if x ≤ −1, then |x + 1| − |x − 1| x = −(x + 1) − −(x − 1) x = −2 x .
You can now sketch the graph by using the fact that the curve y = 1/x is a hyperbola.
4. Exercise 14 in section 2.2: create a table of values for the function and use this to estimate the limit. Then use a graphing utility to graph the function and confirm your result.
Let f (x) = (x3+ 8)/(x + 2). We are to consider the limit of f (x) as x approaches −2. To get a good estimate, you most likely needed to use a calculator. And similarly, it seems like the graph would be difficult to sketch by hand. There is a way to proceed, however.
We can factor (x3+ 8) = (x + 2)(x2− 2x + 4). This is a special case of the following factorization:
(xn+1−yn+1) = (x−y)(xn+xn−1y+xn−2y2+· · ·+x2yn−2+xyn−1+yn). The above can be directly verified by first distributing the x and then the y and then look for cancellation in the resulting sum. Two special cases of the above are when y = 1,
(xn+1− 1) = (x − 1)(xn+ xn−1+ xn−2+ · · · + x2+ x + 1),
and when y = −1,
xn+1+1 = (x+1)(xn−xn−1+xn−2+· · ·+(−1)n−2x2+(−1)n−1x+(−1)n). In our case, we have
x3+ 8 x + 2 =
(x + 2)(x2− 2x + 4)
x + 2 = x
2− 2x + 4
provided that x 6= −2. Since in the limit as x approaches 2, we never let x equal the value of −2, we can say that
lim
x→−2
x3+ 8
x + 2 = limx→−2x
2− 2x + 4 = (−2)2− 2(−2) + 4 = 12.
5. Exercise 42 in section 2.2. Find the limit L of the expression below. Then find δ > 0 such that |f (x) − L| < 0.01 whenever 0 < |x − c| < δ.
lim x→4 4 −x 2
It is clear that the limit will be equal to 4 −42 = 2. So L = 2. Since the limit is as x approaches 4, we see that c = 4.
To solve this exercise, work backwards. We want to be able to conclude that |f (x) − L| < 0.01. Re-writing this, we see that we want to be able to conclude that
|4 − x
2 − 2| = |2 − x
This is what we would like to conclude based on a choice of value of a positive number δ. The role of δ is to specify how close to c = 4 the values of x should be in order for this conclusion to be valid.
For instance if δ = 1, then x can be any number within 1 of c = 4. If, for instance, x = 4.5, then
|f (4.5) − L| = |4 −4.5
2 − 2| = 0.25,
so δ = 1 is not sufficient. We could continue to guess... or we can look for a way to see how the expression |x − c| < δ is related to the expression |f (x) − L|. Since |x − c| = |x − 4|, I will try to factor out this expression: |f (x) − L| = |2 −x 2| = 1 2|x − 4|. Since we want 1 2|x − 4| < 0.01
I see that if |x − 4| < 2(0.01) = 0.02, then this will be true. So, choose δ = 0.02.
