Mathematics 242 – Principles of Analysis Exam 3 Solutions – December 2, 2005

I. f(x) = x^{2} − 6x + 3. A) Let ε > 0 and δ < min{1, ε/5}. If |x − 1| < δ < 1, then 0 < x < 2, so |x − 5| < 5.

Then since we also have |x − 1| < ε/5:

|f(x) − (−2)| = |x^{2} − 6x + 5| = |x − 1||x − 5| < (ε/5) · 5 = ε.

This shows lim_{x→1} f(x) = −2.

B) Note that f′(x) = 2x − 6 ≥ 0 on the interval [3, 4]. Therefore f is monotone increasing. By the result from question C on Discussion 3, f is integrable on that interval. (Alternate way: f is continuous on R, so by the theorem we proved in class on 11/30, f is integrable on every finite interval.)

II. A) IVT: Let f be continuous on [a, b]. If L is any real number (strictly) between f(a)

and f(b), then there exists c ∈ (a, b) such that f(c) = L.

B) f(x) is continuous because f′(x) = −a_{2} sin(x) exists for all real x. Moreover f(0) = a_{1} + a_{2} > 0 while f(π) = a_{1} − a_{2} < 0. Taking L = 0 in the IVT, there exists c ∈ (0, π) such that f(c) = 0.

III. Since s = sup f([a, b]), if ε > 0, then s − ε is not an upper bound for f([a, b]). So there exists x ∈ [a, b] such that s − ε < f(x) ≤ s. Apply this with ε = \frac{1}{n} for each n ∈ N. This gives a sequence (x_{n}) in [a, b] such that s − \frac{1}{n} < f(x_{n}) < s for all n ∈ N. Since [a, b] is a bounded interval, the Bolzano-Weierstrass theorem implies that there is a subsequence x_{n_{k}} of this sequence converging to some c. By the above lim_{k→∞} f(x_{n_{k}}) = s. Since the interval is closed, the limit c is also in the interval [a, b]. But then by the sequential version of continuity

s = lim_{k→∞} f(x_{n_{k}}) = f(c).

IV. A) This is FALSE. Let x_{n} be any sequence of rational numbers converging to 3. Then

lim_{n→∞} f(x) = 9. On the other hand, if y_{n} is a sequence of irrational numbers converging to 3, then lim_{n→∞} f(y_{n}) = 0. Since these two sequential limits are different the limit of the function does not exist.

B) This is TRUE. We have

\frac{f(x) − f(0)}{x − 0} = \{ \begin{matrix}\frac{x^{2}}{x} = x & if x is rational \\ 0 & if x is irrational \end{matrix}

Either way, if |x| < δ = ε, then

|\frac{f(x) − f(0)}{x − 0} − 0| < ε

whenever |x| < ε. So the limit of the difference quotient exists and

f′(0) = lim_{x→0} \frac{f(x) − f(0)}{x − 0} = 0.

C) This is FALSE. If we pick any partition P = {x_{i} : i = 0, …, n} of the interval, then each subinterval contains rational numbers arbitrarily close to the right-hand endpoint of the subinterval. Hence M_{i} = x_{i}^{2}. On the other hand m_{i} = 0 since every subinterval also contains irrational numbers. Since all the x_{i}^{2} > 1, The difference

U(f, P) − L(f, P) = ∑_{i=1}^{n}(M_{i} − m_{i})Δx_{i} = ∑_{i=1}^{n} x_{i}^{2}Δx_{i} > ∑_{i=1}^{n} Δx_{i} = 1.

This shows f is not integrable on [1, 2]. D) This is FALSE. s = sup f([−\sqrt{2}, \sqrt{2}]) = 2 but there is no x in the interval where

f(x) = 2. (Note ±\sqrt{2} are irrational, so f(±\sqrt{2}) = 0).