Fall 2012 Midterm CC-BY-NC

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1Test #1

Held during class on October 4, 2012. 80 minutes (10:05-11:25).

1.1Question 1

(a) State the Archimedean Property for the real numbers.
(b) Given a real number $a$, prove that there is some positive integer $n$ such that $a + n > 0$.


(a) Given any $x \in \mathbb R$, there exists $n \in \mathbb N$ such that $n \geq x$.

(b) Let $x = -a$. Then $x \in \mathbb R$, and by the Archimedean property there exists $n' \in \mathbb N$ such that $n' \geq x$, i.e., $n' \geq -a$. Let $n = n' + 1$. Then $n > n'$, and so $n > -a$; then, if we add $a$ to both sides, we have that $a + n > 0$. $\blacksquare$

1.1.2Accuracy and discussion

See solutions (available through MyCourses)

1.2Question 2

Let $S$ be a non-empty set of real numbers such that every element $x \in S$ satisfies the inequality $x < 2$. For each of the following assertions, state whether or not it is true for all such $S$. If true, give a proof. If not, give an example $S$ for which it is false, and prove that it is false for your example.

(a) $\sup S$ exists (in $\mathbb{R}$).
(b) $\sup S \leq 2$.
(c) $\sup S < 2$.


(a) True. By the completeness property of $\mathbb R$, any non-empty subset in $\mathbb R$ that is bounded above has a supremum. Since 2 is an upper bound for $S$, and it is non-empty by definition, then $S$ has a supremum.

(b) True. We know that 2 is an upper bound, so, by definition of the supremum, we must have that $\sup S \leq 2$.

(c) False. Let $S = \{0 < x < 2\}$. Then, $\sup S = 2$.

1.2.2Accuracy and discussion

See solutions.

1.3Question 3

Let $f$ be the function defined by $f(x) = (x^4-3x+1)/(x^3-5)$ for $2 \leq x \leq 3$.

(a) Prove that the defining expression does actually exist (i.e., that it is defined for all these $x$. (Hint: use inequalities. Do not assume knowledge of numerical values of roots of polynomials.)

(b) Prove that there is a constant $M$ (i.e. a real number) such that for all $x$ in the interval $[2, 3]$ we have $|f(x)| \leq M$.


(a) To prove that it is defined over the given interval, we just need to show that the denominator is never 0. Well, by the triangle inequality, we know that $|x^3-5| \geq ||x^3| - |5|| \geq |x^3| - |5| = |x|^3 - 5$. We also know that $x \geq 2$, so $|x|^3 \geq x^3 \geq 2^3$. Then, we have: $|x^3-5| \geq 2^3-5 = 3$. So the denominator is always $\geq 3$ in this interval, which means that the function is always defined.

(b) First, we find an upper bound for the numerator:

$$\begin{align} |x^4-3x+1| & \leq |x^4| + |-3x| + |1| \tag{by the triangle inequality} \\ & \leq |x^4| + |3x| + 1 \\ & \leq |3^4| + |3\cdot 3| + 1 \tag{as $x \leq 3$} \\ & = 81 + 9 + 1 = 91 \end{align}$$

So 91 is a working upper bound for the numerator. From (a), we have that 3 is a working lower bound for the denominator, which means that

$$\frac{1}{|x^3-5|} \leq \frac{1}{3}$$

in the interval. Combining the two, we get that

$$|f(x)| = \frac{|x^4+3x+1|}{|x^3-5|} \leq \frac{91}{3}$$

Thus we can take $\displaystyle M = \frac{91}{3}$.

1.3.2Accuracy and discussion

See solutions.

2Test #2

Held during class on October 25, 2012. 80 minutes (10:05-11:25).

2.1Question 1

(a) State the definition of $\displaystyle \lim_{n\to \infty} x_n = a$.
(b) Using the definition, prove that
$$\lim_{n\to \infty} \left ( 3n-\sqrt{9n^2-2} \right ) = 0$$


(a) For all $\epsilon > 0$, there exists an $N \in \mathbb N$ such that for all $n \geq N$, $|x_n - a| < \epsilon$.

(b) First, we try to get the expression $|x_n-a$ in a nicer form, in order to find an upper bound for it:

$$\begin{align} |x_n-a| & = |(3n-\sqrt{9n^2-2}) - 0| = |3n - \sqrt{9n^2-2}| \\ & = \left | (3n - \sqrt{9n^2-2}) \cdot \frac{3n + \sqrt{9n^2-2}}{3n + \sqrt{9n^2-2}} \right | \tag{multiplying by the conjugate} \\ & = \left | \frac{9n^2 - (9n^2-2)}{3n + \sqrt{9n^2-2}} \right | = \left | \frac{2}{3n+\sqrt{9n^2-2}} \right | \\ & < \left | \frac{2}{3n + 0} \right | \tag{since $\sqrt{9n^2-2} > 0$ for $n > 0$} \\ & = \frac{2}{3n} \tag{since $n > 0$} \end{align}$$

Then, given some $\epsilon > 0$, let $N = \frac{3}{2\epsilon}$. For all $n \geq N$, we have:

$$|x_n - a| < \frac{2}{3N} = \frac{2}{3(\frac{3}{2\epsilon})} = \epsilon$$

which shows that $|x_n-a| < \epsilon$ for $a = 0$, proving that the limit is 0.

2.1.2Accuracy and discussion

See solutions on MyCourses. My answer for (b) differs slightly from the provided solution (I used the fact that $\sqrt{9n^2-2} > 0$ for all $n \in \mathbb N$, which I think I proved by induction or something - at any rate, it's trivial). I got full marks for it though. - @dellsystem

2.2Question 2

Let $x_1 = 1$ and $x_{n+1} = \sqrt{6+x_n}$ for $n \in \mathbb N$.

(a) Prove that the sequence $(x_n)$ converges.
(b) Find $\displaystyle \lim_{n\to \infty} x_n$.


(a) We can use the Monotone Convergence Theorem: if a sequence is either bounded above and increasing or bounded below and decreasing, then it converges. For this particular sequence, we will prove that it is bounded above and increasing.

We will first show that it is increasing, i.e., $x_{n+1} \geq x_{n}$, by induction. In the base case, $x_1 = 1$, and $x_2 = \sqrt{6 + x_1} = \sqrt{7}$. Clearly $\sqrt{7} \geq 1$, so $x_2 \geq x_1$ $\checkmark$. For the induction hypothesis, we assume that $x_{n+1} \geq x_n$. Then, if we add 6 to both sides, we have that $6 + x_{n+1} \geq 6 + x_n$. Taking the square root of both sides gives $\sqrt{6 + x_{n+1}} = x_{n+2} \geq \sqrt{6 + x_n} = x_{n+1}$ which tells us that $x_{n+2} \geq x_{n+1}$, and so the induction hypothesis holds for $n+1$.

Now we will show that it is bounded above. We posit that 3 is an upper bound for the sequence. (In fact it is actually the least upper bound, though that'll come later.) We again use a proof by induction. In the base case, $x_1 = 1 \leq 3$ $\checkmark$. For the induction hypothesis, we assume that $x_n \leq 3$. Then, if we add 6 to both sides, we get that $6 + x_n \leq 6 + 3 = 9$, and taking the square root of both sides gives us $\sqrt{6 + x_n} \leq \sqrt{9} = 3$. So $\sqrt{6+x_n} = x_{n+1} \leq 3$ which completes this proof by induction. $\checkmark$

Consequently, by the Monotone Convergence Theorem, this sequence converges.

(b) Let $x$ be the limit. By limit theorems, we know that $x = \sqrt{6 + x}$. Squaring both sides and moving stuff around gives us $x^2 - x - 6 = 0$. We can factor that as $(x-3)(x+2) = 0$, meaning that $x$ is either 3 or -2. Since the first term is 1, and the sequence is increasing, we know that none of the terms can be less than 1 and so -2 is not a possible limit. So $x = 3$.

2.2.2Accuracy and discussion

The provided solution for (b) uses the quadratic formula, but I never liked the quadratic formula

2.3Question 3

Given the existence of the limit $\displaystyle \lim_{n\to\infty} \left (1 + \frac{1}{n} \right )^n = e$, prove that the sequence

$$x_n = \left ( \frac{1}{1+\frac{1}{n}} \right )^{2n+1}, \quad n \in \mathbb N$$

converges and find its limit. (You may use any limit theorems covered in class.)


We can rewrite $x_n$ in terms of the other sequence as follows:

$$x_n = \left ( \frac{1}{1+\frac{1}{n}} \right )^{2n+1} = \frac{1}{\left ( 1 + \frac{1}{n} \right)^{2n+1}} = \frac{1}{\left [\left (1+\frac{1}{n} \right )^n \right ]^2 \cdot \left ( 1 + \frac{1}{n} \right )}$$

Now, we know that $1 + \frac{1}{n}$ converges to 1, and the other thing in the denominator converges to $e^2$, so the denominator converges to something non-zero. Consequently, the sequence $x_n$ converges, and its limit is

$$\lim x_n = \frac{1}{e^2\cdot 1} = \frac{1}{e^2}$$

(using limit theorems for quotients and products).

2.3.2Accuracy and discussion

See solutions.

3Test #3

3.1Question 1

Let $x_n = \sqrt{n}$. Prove directly from the definition that $(x_n)$ is not a Cauchy sequence.


Let $n = N$ and $m = 4N$ where $N \in \mathbb N$. Then $| \sqrt{n} - \sqrt{m}| = |\sqrt{N} - \sqrt{4N}| = |\sqrt{N} - 2\sqrt{N}| = |-\sqrt{N}| = \sqrt{N} \geq 1$ (since $N \geq 1$). Then, if $\epsilon = 1$, there does not exist an $N$ such that $|x_m - x_n| < \epsilon$ for all $m > n > N$, which proves that this sequence is not a Cauchy sequence.

3.1.2Accuracy and discussion

See solutions.

3.2Question 2

Using the $\epsilon$-$\delta$ definition of limit, prove that $\displaystyle \lim_{x \to 3}x^3 = 27$.


First, we rewrite $|x^3-27|$ to find an upper bound for it:

$$\begin{align}|x^3-27| & = |x^3-3^3| = |x-3||x^2+3x+9| \\ & \leq |x-3|(|x|^2+3|x| + 9|) \tag{by the triangle inequality} \\ & \leq |x-3|(4^2 + 3\cdot 4 + 9) \tag{provided $|x-3| \leq 1$} \\ & = |x-3| \cdot 37 \end{align}$$

Then, for a given $\epsilon > 0$, let $\delta = \inf\{1, \frac{\epsilon}{37}\}$. Then, by the above work, if $|x-3| < \delta$, then $|x^3-27| < \epsilon$ which proves that the limit of $x^3$ at $x=3$ is 27.

3.2.2Accuracy and discussion

3.3Question 3

Let $a, L \in \mathbb R$ and let $f: \mathbb R \to \mathbb R$ be such that $\displaystyle \lim_{x \to a} (f(x)f(x)) = L$.

(a) Suppose that $L=0$. Prove that $\displaystyle \lim_{x\to a} f(x) = 0$, using the $\epsilon$-$\delta$ definition of limit.
(b) Suppose that $L = 1$, $a = 0$. Show by example that $f$ may not have a limit at $a$.


(a) By the definition of limit, we know that if $(f(x)f(x))$ has a limit of 0 at $a$, then there exists a $\delta > 0$ such that for any $\epsilon > 0$, we have $|f(x)f(x)| < \epsilon$ whenever $0 < |x-a| < \delta$. Now, since $\epsilon^2 > 0$ for any $\epsilon > 0$, we know there exists a $\delta$ such that $0 < |x-a| < \delta$ results in $|f(x)^2| < \epsilon^2$. We could also write that as $|f(x)|^2 < \epsilon^2$. Taking the square root of both sides gives us that $|f(x)| < \epsilon$ whenever $0 < |x-a| < \delta$ which tells us that 0 is the limit of $f(x)$ as well.

(b) We refer to the signum function. Then $(f(x))^2 = 1$ for all $x \neq 0$. So $\lim_{x \to 0} (f(x))^2 = 1$. But $\lim_{x \to 0} f(x)$ does not exist since it's 1 for positive $x$ and -1 for negative $x$.

3.3.2Accuracy and discussion

See solutions.