Mathematics 242 – Principles of Analysis Exam 3 – Solutions December 6, 2004

I. A) Let f : D → R and let c be an accumulation point of D. We say lim_{x→c} f(x) = L if for all ε > 0, there exists δ > 0 such that |f(x) − L| < ε for all x ∈ D with 0 < |x − c| < δ.

B) The domain of f is D = R in this case. Given ε > 0, let δ = \sqrt{ε}. Then for all x with 0 < |x − 2| < \sqrt{ε},

|f(x) − (−2)| = |x^{2} − 4x + 4| = |x − 2|^{2} < (\sqrt{ε})^{2} = ε.

Hence the definition is satisfied and lim_{x→2} x^{2} − 4x + 2 = −2.

C) The limit lim_{x→0} \frac{x^{2}}{sin(2x^{2})} is indeterminate of the form 0/0 and the other hypotheses for L’Hopital’s Rule are satisfied, so we apply L’Hopital to see:

lim_{x→0} \frac{x^{2}}{sin(2x^{2})} = lim_{x→0} \frac{2x}{4x cos(2x^{2})}

= lim_{x→0} \frac{1}{2 cos(2x^{2})}

= 1/2

(Note: The limit on the right side of the first line here is also indeterminate 0/0. So it could be evaluated by applying L’Hopital a second time. But canceling the x’s is certainly easier!)

II. A) Intermediate Value Theorem. Let f be continuous on [a, b]. (Note: no differentiability hypotheses are necessary here!) Assume f(a) ≠ f(b). Then for all y between f(a) and f(b), there exists at least one x ∈ [a, b] such that f(x) = y.

B) f(x) = a_{1} + a_{2} sin(x) is continuous for all x ∈ R (since f is differentiable, for instance!) So the IVT applies on the interval [−π/2, π/2]. We have f(−π/2) = a_{1} + a_{2} sin(−π/2) = a_{1} − a_{2} < 0 since a_{2} > a_{1}. On the other hand f(π/2) = a_{1} + a_{2} > 0 since a_{1}, a_{2} > 0. Hence f changes sign on this interval and by the IVT, there exists c ∈ [−π/2, π/2] such that f(c) = 0.

III. Here’s the proof we did in class. (This is slightly different from the one in the text; Lay doesn’t assume the partition is regular, but argues that if all the smaller intervals in the partition are sufficiently small, then the difference between the upper and lower sums can be made arbitrarily small. Both proofs are correct of course!). Let P_{n} = {a = x_{0}, x_{1}, …, x_{n−1}, x_{n} = b} be the regular partition of [a, b] with Δx = (b − a)/n. Then since f is monotone increasing,

U(f, P_{n}) = ∑_{i=1}^{n} M_{i}Δx = ∑_{i=1}^{n} f(x_{i})Δx

and

L(f, P_{n}) = ∑_{i=1}^{n} m_{i}Δx = ∑_{i=1}^{n} f(x_{i−1})Δx

In the difference, U(f, P_{n}) − L(f, P_{n}), the sum “telescopes” and we get

U(f, P_{n}) − L(f, P_{n}) = (f(b) − f(a))Δx = (f(b) − f(a))(b − a)/n

Given any ε > 0, take n > (f(b) − f(a))(b − a)/ε. Then U(f, P_{n}) − L(f, P_{n}) < ε. (This shows that f is integrable by the result of Exercise 29.9 on the last problem set – the idea is that since U(f, P_{n}) and L(f, P_{n}) can be made arbitrarily close by taking n large enough, U(f) = inf_{P} {U(f, P)} and L(f) = sup_{P} {L(f, P)} must be equal.

IV. A) This is FALSE. f is not continuous at 0 (because lim_{x→0^{−}} f(x) = −2, while lim_{x→0^{+}} f(x) = +2). So f is not differentiable at 0 either. This can also be seen by trying to compute the limit of the difference quotient for f. If x ≥ 0, f(x) = 2 − x − x^{3}, so

lim_{x→0^{+}} \frac{f(x) − f(0)}{x − 0} = lim_{x→0^{+}} \frac{−x − x^{2}}{x} = lim_{x→0^{+}} −1 − x = −1.

But

lim_{x→0^{−}} \frac{f(x) − f(0)}{x − 0} = lim_{x→0^{−}} \frac{−x − 4}{x} = lim_{x→0^{−}} −1 − \frac{4}{x},

which does not exist (we might say the limit goes to +∞). Hence f′(0) does not exist.

B) This is TRUE. Full credit was given for the following argument: Split the integral at x = 0. Since f is continuous and monotone decreasing on [−2, 0] and [0, 2], then we know f is integrable on each half by the monotone decreasing case of the theorem from question III. Then

∫_{−2}^{2} f(x) dx = ∫_{−2}^{0} f(x) dx + ∫_{0}^{2} f(x) dx

exists too. (You could also get the same result by considering partitions containing x_{k} = 0 for some k, so the upper and lower sums for f on the whole interval split into upper and lower sums for f on each half. Since f is integrable on each half of the interval, the upper and lower sums can be made arbitrarily close.) This reasoning suggests an important general statement: If f is continuous except at a finite number of points in [a, b], or if [a, b] can be split into a finite number of intervals on which f is monotonic, then f is integrable on [a, b]. This can be proved in the same way – split into subintervals, and apply the appropriate theorem on each piece.

C) This is TRUE. The conclusion of the MVT is that there exists some c in the interval [a, b] such that f′(c) = \frac{f(b)−f(a)}{b−a}. The hypotheses of the MVT are not satisfied for this function on the interval [−2, 2] (see part A). HOWEVER, note that \frac{f(2)−f(−2)}{2−(−2)} = \frac{−8−0}{4} = −2. Moreover, f is differentiable for all x ≠ 0, and

f′(x) = \{ \begin{matrix}−1 & if x < 0 \\ −1 − 3x^{2} & if x > 0 \end{matrix}

We have −2 = −1 − 3x^{2} if 3x^{2} = 1, or x = ±1/\sqrt{3}. x = 1/\sqrt{3} is in our interval, so the conclusion of of the MVT is satisfied!

D) This is FALSE. The conclusion of the EVT is that, writing M = sup{f(x) : x ∈ [a, b]} and m = inf{f(x) : x ∈ [a, b]}, there exist c_{1}, c_{2} ∈ [a, b] such that f(c_{1}) = M and f(c_{2}) = m. Again, the hypotheses of the EVT are not satisfied on [−2, 1] (because f is not continous at 0). Here, by considering the graph of f, we can see M = 2 = f(0), so f does attain its maximum on the interval. But, m = −2 since lim_{x→0^{−}} f(x) = −2. But there is no x ∈ [−2, 1] such that f(x) = −2 (note f(0) = +2, not −2 by the definition of f(x), and the smallest value of f for x between 0 and 1 is f(1) = 0.)

General comment on parts C, D: The point of these questions was not only to test your knowledge of the relevant theorems, but also to test your understanding of the underlying logic of mathematical statements. Even if the hypotheses of a theorem are not satisfied, it is possible for the conclusion to be true (see part C). However, there is no guarantee that the conclusion must be true (see part D).