Mathematics 242 – Principles of Analysis Exam 2 – Solutions

I. A) See the class notes or the text. B) We want to show that lim_{x→2} x^{2} − 5x = −6. First note that

x^{2} − 5x − (−6) = (x − 2)(x − 3), so |x^{2} − 5x + 6| = |x − 2||x − 3|

If 0 < |x − 2| < δ, and δ < 1, then 1 < x < 3, so −2 < x − 3 < 0. Hence |x − 3| < 2. Therefore, if ϵ > 0 is any given positive number, δ < min(1, ϵ/2), and 0 < |x − 2| < δ, then we have

|x^{2} − 5x + 6| = |x − 2||x − 3|
< (ϵ/2) · 2
= ϵ

II. See the class notes (or the notes from the review session, or the text).

III. A) The answer is no. We have

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

provided that the limit exists. But if we let x → 0^{−}, then since f(x) = 1 for all x < 0, we see

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

On the other hand letting x → 0^{+}, we have

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

Since the two one-sided limits are not the same, f′(0) does not exist. B) Full credit was given for the answer “f is integrable, since it is continuous on [−1, 1].”

This can also be proved directly without invoking that theorem, by considering partitions as indicated in the problem statement. Note that f is strictly increasing – not constant on the interval from 0 to 1, so we will not be able to get the difference between an upper and lower sum to be arbitrarily small by picking partitions with only one or two subintervals. Instead, we note that if P_{n} = {−1, x_{1} = 0, x_{2}, …, x_{n} = 1} where P′ = {0, x_{2}, …, 1} is a regular partition of [0, 1] (with n − 1 smaller intervals), then since f(x) is strictly increasing on [0, 1] we will have:

U_{P_{n}}(f) = 1 · 1 + ∑_{i=2}^{n} f(x_{i})Δx

L_{P_{n}}(f) = 1 · 1 + ∑_{i=2}^{n} f(x_{i−1})Δx, so
U_{P_{n}}(f) − L_{P_{n}}(f) = (f(1) − f(0))Δx
= 2Δx

This can be made arbitrarily small by picking Δx small enough (that is taking n = \frac{1}{Δx} + 1 large enough). Hence f is integrable on [−1, 1].

C) The value of the integral ∫_{−1}^{1} f(x) dx may be found without the Fundamental Theorem of Calculus by taking the limit of the upper sums for a partition as in part B. For simplicity we write m = n − 1 (the number of smaller intervals after the first, “big” one from −1 to 0):

∫_{−1}^{1} f(x) dx = lim_{n→∞} U_{P_{n}}(f)

= lim_{m→∞} 1 + ∑_{i=1}^{m} (1 + \frac{i}{m} + (\frac{i}{m})^{2}) \frac{1}{m}

= 1 + lim_{m→∞} (\frac{1}{m} ∑_{i=1}^{m} 1 + \frac{1}{m^{2}} ∑_{i=1}^{m} i + \frac{1}{m^{3}} ∑_{i=1}^{m} i^{2})

= 1 + lim_{m→∞} (1 + \frac{1}{m^{2}} \frac{m(m + 1)}{2} + \frac{1}{m^{3}} \frac{m(m + 1)(2m + 1)}{6})

= 1 + 1 + \frac{1}{2} + \frac{1}{3}

= \frac{17}{6}

IV. True-False.

A) This statement is FALSE:

\{ \begin{matrix}lim_{x→0^{−}} \frac{sin(x)}{|x|} = −1 & but \\ lim_{x→0^{+}} \frac{sin(x)}{|x|} = +1 \end{matrix}

Therefore, \frac{sin(x)}{|x|} has a jump discontinuity at x = 0, not a removable discontinuity.

B) This statement is also FALSE. The IVT has no converse statement. For example the

function:

f(x) = \{ \begin{matrix}x^{2} & if 0 ≤ x ≤ 1 \\ \frac{1}{2} & if 1 < x ≤ 2 \end{matrix}

has the intermediate value property on [0, 2] – for every y between y = m = 0 and y = M = 1, there is some x such that f(x) = y. But f is not continuous on [0, 2] because of the jump at 1.

C) This statement is TRUE. By hypothesis, f′′(x) exists for all x. This implies that f′(x) exists and is continuous for all x. Similarly, f(x) exists and is continuous for all x. We may assume x_{1} < x_{2} < x_{3}. By Rolle’s Theorem (MVT) on [x_{1}, x_{2}] there is some c_{1} ∈ (x_{1}, x_{2}) such that f′(c_{1}) = 0. Similarly, on [x_{2}, x_{3}], there is some c_{2} such that f′(c_{2}) = 0. But now by Rolle’s Theorem applied to f′(x) on the interval [c_{1}, c_{2}], there is some d ∈ (c_{2}, c_{2}) such that (f′)′(d) = f′′(d) = 0.