§8 A Discussion about Proofs

This section works through several proofs that use the definition of the limit directly. With some study and practice you should be able to write proofs like these yourself. Each example has a discussion, which is the scratch work you would do on paper, followed by the formal proof you would hand in.

Intuitionthe shape of every ε–N proof

Every proof that $s_n \to s$ has the same skeleton:

Let $\varepsilon > 0$. Let $N = \underline{\quad\quad}$. Then $n > N$ implies $\cdots$, so $|s_n - s| < \varepsilon$.

The scratch work is a search for what to put in the blank, and it usually runs backward from the goal. The formal proof runs forward and only has to check that the chosen $N$ works. A reader should never have to see how you found it.

Example 1.

Prove $\lim \dfrac{1}{n^2} = 0$.

Discussion.

We must take an arbitrary $\varepsilon > 0$ and produce an $N$, depending on $\varepsilon$, such that $n > N$ implies $\left|\frac1{n^2} - 0\right| < \varepsilon$. So the formal proof should begin “Let $\varepsilon > 0$” and end with “Hence $n > N$ implies $\left|\frac1{n^2} - 0\right| < \varepsilon$.” In between, it must name $N$ and check that it works.

To find $N$, we work backward from what we want, as you might with a trigonometric identity, making sure every step is reversible. We want $\frac1{n^2} < \varepsilon$. Multiplying by $n^2$ and dividing by $\varepsilon$, this is $\frac1\varepsilon < n^2$, or $\frac1{\sqrt\varepsilon} < n$. This suggests $N = \frac1{\sqrt\varepsilon}$.

Formal Proof

Let $\varepsilon > 0$ and let $N = \frac1{\sqrt\varepsilon}$. Then $n > N$ implies $n > \frac1{\sqrt\varepsilon}$, which implies $n^2 > \frac1\varepsilon$ and hence $\varepsilon > \frac1{n^2}$. Thus $n > N$ implies $\left|\frac1{n^2} - 0\right| < \varepsilon$. This proves $\lim \frac1{n^2} = 0$.

∎

Example 2.

Prove $\lim \dfrac{3n+1}{7n-4} = \dfrac37$.

Discussion.

For each $\varepsilon > 0$ we must decide how big $n$ must be to guarantee $\left|\frac{3n+1}{7n-4} - \frac37\right| < \varepsilon$. Combining the fractions, we want

$$\left| \frac{21n + 7 - 21n + 12}{7(7n - 4)} \right| < \varepsilon \qquad\text{or}\qquad \frac{19}{7(7n-4)} < \varepsilon.$$

Since $7n - 4 > 0$ we dropped the absolute value. Now solve for $n$:

$$\frac{19}{7\varepsilon} < 7n - 4 \qquad\text{or}\qquad \frac{19}{7\varepsilon} + 4 < 7n \qquad\text{or}\qquad \frac{19}{49\varepsilon} + \frac47 < n.$$

These steps are reversible, so we put $N = \frac{19}{49\varepsilon} + \frac47$. Any larger $N$ would also work.

Formal Proof

Let $\varepsilon > 0$ and let $N = \frac{19}{49\varepsilon} + \frac47$. Then $n > N$ implies $n > \frac{19}{49\varepsilon} + \frac47$, hence $7n > \frac{19}{7\varepsilon} + 4$, hence $7n - 4 > \frac{19}{7\varepsilon}$, hence $\frac{19}{7(7n-4)} < \varepsilon$, and hence $\left|\frac{3n+1}{7n-4} - \frac37\right| < \varepsilon$. This proves the claim.

∎

This is the $N$ that the ε–N widget in §7 uses for this sequence.

Example 3.

Prove $\lim \dfrac{4n^3 + 3n}{n^3 - 6} = 4$.

Discussion.

We must find how large $n$ has to be to make

$$\left| \frac{4n^3 + 3n}{n^3 - 6} - 4 \right| < \varepsilon \qquad\text{or}\qquad \left| \frac{3n + 24}{n^3 - 6} \right| < \varepsilon.$$

For $n > 1$ we can drop the absolute values. This time it would be very hard to solve for $n$. But we do not need the least $N$, only some $N$, so we simplify with estimates. The idea is that $\frac{3n+24}{n^3-6}$ is at most a constant times $\frac{n}{n^3} = \frac1{n^2}$ for large $n$. To get such a bound we make the numerator bigger and the denominator smaller.

Since $3n + 24 \le 27n$, it suffices to get $\frac{27n}{n^3 - 6} < \varepsilon$. Next, $n^3 - 6 \ge \frac{n^3}{2}$ once $\frac{n^3}2 \ge 6$, which holds for $n > 2$. So for $n > 2$ it suffices to get $\frac{27n}{n^3/2} < \varepsilon$, i.e. $\frac{54}{n^2} < \varepsilon$, i.e. $n > \sqrt{54/\varepsilon}$.

Formal Proof

Let $\varepsilon > 0$ and let $N = \max\left\{2, \sqrt{54/\varepsilon}\right\}$. Then $n > N$ implies $n > \sqrt{54/\varepsilon}$, hence $\frac{54}{n^2} < \varepsilon$, hence $\frac{27n}{n^3/2} < \varepsilon$. Since $n > 2$, we also have $\frac{n^3}2 \le n^3 - 6$, and $27n \ge 3n + 24$. So $n > N$ implies

$$\frac{3n + 24}{n^3 - 6} \le \frac{27n}{\frac12 n^3} = \frac{54}{n^2} < \varepsilon, \qquad\text{hence}\qquad \left| \frac{4n^3 + 3n}{n^3 - 6} - 4 \right| < \varepsilon.$$∎

Intuitionhow to estimate

You want $A_n < \varepsilon$, but $A_n$ is too messy to solve for $n$. Find a simpler $B_n$ with $A_n \le B_n$ for all large $n$, and make $B_n < \varepsilon$ instead. To make a fraction bigger, increase the numerator or decrease the denominator, never the reverse. Each simplification usually needs $n$ to be large enough, like $n > 2$ above. Collect those requirements and build them into $N$ with a $\max$.

Common pitfall

• When shrinking a denominator, keep it positive. $n^3 - 6 \ge \frac{n^3}{2}$ is useful only because both sides are positive once $n > 2$.
• Backward scratch work is safe only if each step can be reversed. If one cannot, check the forward direction separately in the formal proof.
• Every assumption about $n$ has to be justified by $n > N$. If your estimate needs $n \ge 10$, the formal proof needs $N \ge 10$.

As Example 3 shows, direct proofs of even simple limits can get messy. With the limit theorems of §9 we will just write

$$\lim \frac{4n^3 + 3n}{n^3 - 6} = \lim \frac{4 + \frac3{n^2}}{1 - \frac6{n^3}} = \frac{4 + 3\lim\frac1{n^2}}{1 - 6\lim\frac1{n^3}} = 4.$$

Example 4.

Show that the sequence $a_n = (-1)^n$ does not converge.

Discussion.

Assume $\lim (-1)^n = a$ and look for a contradiction. Whatever $a$ is, at least one of $1$ and $-1$ is at distance at least $1$ from $a$. So $|(-1)^n - a| < 1$ cannot hold for all large $n$.

Formal Proof

Assume $\lim(-1)^n = a$ for some $a \in \mathbb{R}$. Taking $\varepsilon = 1$ in the definition of limit, there is $N$ such that $n > N$ implies $|(-1)^n - a| < 1$. Using an even and an odd $n > N$, we get $|1 - a| < 1$ and $|-1 - a| < 1$. By the Triangle Inequality 3.7,

$$2 = |1 - (-1)| = |1 - a + a - (-1)| \le |1 - a| + |a - (-1)| < 1 + 1 = 2.$$

This absurdity shows the assumption was wrong, so $(-1)^n$ does not converge.

∎

Intuitionproving divergence

To show $s_n \not\to a$, negate Definition 7.1: find one $\varepsilon > 0$ such that, however large $N$ is, some $n > N$ has $|s_n - a| \ge \varepsilon$. Example 4 does this with $\varepsilon = 1$ for every candidate $a$ at once. A single bad tolerance is enough.

Example 5.

Let $(s_n)$ be a sequence of nonnegative real numbers with $s = \lim s_n$. Note $s \ge 0$ (Exercise 8.9(a)). Prove $\lim \sqrt{s_n} = \sqrt{s}$.

Discussion.

This time we cannot expect a formula for $N$ in terms of $\varepsilon$, because the problem is so general. But we can show some $N$ exists. The trick is to break with algebra training and “irrationalize the denominator”:

$$\sqrt{s_n} - \sqrt{s} = \frac{(\sqrt{s_n} - \sqrt s)(\sqrt{s_n} + \sqrt s)}{\sqrt{s_n} + \sqrt s} = \frac{s_n - s}{\sqrt{s_n} + \sqrt s}.$$

Since $s_n \to s$ we can make the numerator small. If $s > 0$ the denominator is at least $\sqrt s$, so $|\sqrt{s_n} - \sqrt s| \le \frac{|s_n - s|}{\sqrt s}$, and we choose $N$ so that $|s_n - s| < \sqrt s\,\varepsilon$ for $n > N$. Such an $N$ exists because the definition of limit can be applied to $\sqrt s\,\varepsilon$ just as well as to $\varepsilon$. If $s = 0$ the denominator can be small too, but then a direct argument works (Exercise 8.3).

Formal Proof

Case I: $s > 0$. Let $\varepsilon > 0$. Since $\lim s_n = s$, there is $N$ such that $n > N$ implies $|s_n - s| < \sqrt s\,\varepsilon$. Then $n > N$ implies

$$|\sqrt{s_n} - \sqrt s| = \frac{|s_n - s|}{\sqrt{s_n} + \sqrt s} \le \frac{|s_n - s|}{\sqrt s} < \frac{\sqrt s\,\varepsilon}{\sqrt s} = \varepsilon.$$

Case II: $s = 0$. This is Exercise 8.3.

∎

Example 6.

Let $(s_n)$ be a convergent sequence with $s_n \ne 0$ for all $n$ and $\lim s_n = s \ne 0$. Prove $\inf\{|s_n| : n \in \mathbb{N}\} > 0$.

Discussion.

“Most” terms $s_n$ are close to $s$ and hence not close to $0$. More precisely, most $s_n$ are within $\frac12|s|$ of $s$, so most satisfy $|s_n| \ge \frac12 |s|$. The finitely many remaining terms are nonzero, so they cannot drag the infimum down to $0$.

Formal Proof

Let $\varepsilon = \frac12|s| > 0$. Since $\lim s_n = s$, there is $N \in \mathbb{N}$ such that $n > N$ implies $|s_n - s| < \frac{|s|}2$. Then

$$n > N \quad\text{implies}\quad |s_n| \ge \frac{|s|}{2}, \tag{1}$$

since otherwise the Triangle Inequality would give $|s| = |s - s_n + s_n| \le |s - s_n| + |s_n| < \frac{|s|}2 + \frac{|s|}2 = |s|$, which is absurd. Put $m = \min\left\{\frac{|s|}2, |s_1|, |s_2|, \dots, |s_N|\right\}$. Then $m > 0$ and, by (1), $|s_n| \ge m$ for all $n$. So $\inf\{|s_n| : n \in \mathbb{N}\} \ge m > 0$.

∎

Reading alongside Abbott

Abbott’s Section 2.2 works through similar examples and stresses the same two stages: rough work to find $N$, then a clean forward proof that starts with “Let $\varepsilon > 0$”.

Practice: spot the error

This is the classic slip in turning scratch work into a proof.

Spot the error: the limit of 1/n

One line of this argument is wrong. Press the button on the line you think is the mistake.

1. Claim: $\lim \frac1n = 0$. Let $\varepsilon > 0$.
2. Suppose that $\left|\frac1n - 0\right| < \varepsilon$.
3. Then $\frac1n < \varepsilon$, which means $n > \frac1\varepsilon$.
4. So take $N = \frac1\varepsilon$. Hence $\lim \frac1n = 0$.

Check your understanding

Take the quiz (3 questions)

To prove $\lim \frac{1}{n^2} = 0$, given $\varepsilon > 0$, which choice of $N$ works?

$N = \frac{1}{\sqrt\varepsilon}$$N = \varepsilon$$N = \sqrt\varepsilon$

Your scratch work shows $|s_n - s| < \varepsilon$ whenever $n > \frac3\varepsilon$. How should the formal proof begin?

”Let $\varepsilon > 0$ and let $N = \frac3\varepsilon$. Then $n > N$ implies …""Let $n > N$. Choose $\varepsilon$ so that …""Assume $|s_n - s| < \varepsilon$. Then …”

If $\lim s_n = s$ and $s > 0$, which follows?

$s_n > 0$ for all $n$.There is $N$ such that $s_n > \frac s2$ for all $n > N$.$s_n \ge s$ for all large $n$.

Exercises

Formal proofs are required in the following exercises.

Exercise 8.1

Prove the following:

1. $\lim \dfrac{(-1)^n}{n} = 0$
2. $\lim \dfrac{1}{n^{1/3}} = 0$
3. $\lim \dfrac{2n-1}{3n+2} = \dfrac23$
4. $\lim \dfrac{n+6}{n^2-6} = 0$

Hint

(a) $N = \frac1\varepsilon$. (b) $N = \frac1{\varepsilon^3}$. (c) $\left|\frac{2n-1}{3n+2} - \frac23\right| = \frac{7}{3(3n+2)} < \frac{7}{9n}$, so $N = \frac{7}{9\varepsilon}$ works. (d) For $n \ge 6$: $n + 6 \le 2n$ and $n^2 - 6 \ge \frac{n^2}2$, so the term is at most $\frac4n$. Take $N = \max\{6, \frac4\varepsilon\}$.

Solution

In each part let $\varepsilon > 0$ and find $N$ so that $n > N$ implies $|s_n - s| < \varepsilon$.

1. $\left|\frac{(-1)^n}{n} - 0\right| = \frac1n < \varepsilon$ whenever $n > N = \frac1\varepsilon$.
2. $\frac1{n^{1/3}} < \varepsilon$ iff $n^{1/3} > \frac1\varepsilon$ iff $n > \frac1{\varepsilon^3}$; take $N = \frac1{\varepsilon^3}$.
3. $\dfrac{2n-1}{3n+2} - \dfrac23 = \dfrac{3(2n-1) - 2(3n+2)}{3(3n+2)} = \dfrac{-7}{3(3n+2)}$, so the distance is $\dfrac{7}{3(3n+2)} < \dfrac7{9n}$. This is $< \varepsilon$ when $n > N = \frac7{9\varepsilon}$.
4. For $n \ge 6$ we have $n + 6 \le 2n$ and $n^2 - 6 \ge \frac{n^2}2$ (since $n^2 \ge 12$), so $0 < \frac{n+6}{n^2-6} \le \frac{2n}{n^2/2} = \frac4n$. This is $< \varepsilon$ for $n > N = \max\{6, \frac4\varepsilon\}$.

Tags: Standardε–N / ε–δ

Exercise 8.2

Determine the limits of the following sequences, and then prove your claims.

1. $a_n = \dfrac{n}{n^2+1}$
2. $b_n = \dfrac{7n-19}{3n+7}$
3. $c_n = \dfrac{4n+3}{7n-5}$
4. $d_n = \dfrac{2n+4}{5n+2}$
5. $s_n = \dfrac1n \sin n$

Solution

1. $0$
2. $\tfrac73$
3. $\tfrac47$
4. $\tfrac25$
5. $0$

For (b), for example: $\left|\frac{7n-19}{3n+7} - \frac73\right| = \frac{106}{3(3n+7)} < \frac{106}{9n}$, so take $N = \frac{106}{9\varepsilon}$. For (e), $\left|\frac{\sin n}{n}\right| \le \frac1n$.

Tags: Standardε–N / ε–δ

Exercise 8.3

Let $(s_n)$ be a sequence of nonnegative real numbers with $\lim s_n = 0$. Prove $\lim \sqrt{s_n} = 0$. This completes the proof of Example 5.

Hint

Given $\varepsilon > 0$, apply the definition of $\lim s_n = 0$ with $\varepsilon^2$ in place of $\varepsilon$.

Solution

Let $\varepsilon > 0$. Since $s_n \to 0$, applying the definition with $\varepsilon^2$ gives $N$ such that $n > N$ implies $s_n < \varepsilon^2$ (recall $s_n \ge 0$). Then $0 \le \sqrt{s_n} < \varepsilon$ for $n > N$, so $\sqrt{s_n} \to 0$.

Tags: Standardε–N / ε–δ

Exercise 8.4

Let $(t_n)$ be a bounded sequence, i.e. there is $M$ with $|t_n| \le M$ for all $n$, and let $(s_n)$ be a sequence with $\lim s_n = 0$. Prove $\lim (s_n t_n) = 0$.

Hint

We may take $M > 0$. Choose $N$ so that $|s_n| < \frac{\varepsilon}{M}$ for $n > N$.

Solution

We may assume $M > 0$. Let $\varepsilon > 0$ and choose $N$ so that $|s_n| < \frac\varepsilon M$ for $n > N$. Then for $n > N$, $|s_nt_n - 0| = |s_n||t_n| \le M|s_n| < \varepsilon .$

Tags: Standardε–N / ε–δ

Exercise 8.5

This exercise is referred to in several places.

1. Consider sequences $(a_n)$, $(b_n)$, $(s_n)$ with $a_n \le s_n \le b_n$ for all $n$ and $\lim a_n = \lim b_n = s$. Prove $\lim s_n = s$. This is the squeeze lemma.
2. Suppose $|s_n| \le t_n$ for all $n$ and $\lim t_n = 0$. Prove $\lim s_n = 0$.

Hint

(a) For $n$ beyond both $N$’s, $s - \varepsilon < a_n \le s_n \le b_n < s + \varepsilon$. (b) Apply (a) with $a_n = -t_n$ and $b_n = t_n$.

Solution

1. Let $\varepsilon > 0$. Choose $N_1$ with $|a_n - s| < \varepsilon$ for $n > N_1$ and $N_2$ with $|b_n - s| < \varepsilon$ for $n > N_2$. For $n > \max\{N_1, N_2\}$, $s - \varepsilon < a_n \le s_n \le b_n < s + \varepsilon,$ so $|s_n - s| < \varepsilon$.
2. From $|s_n| \le t_n$ we get $-t_n \le s_n \le t_n$. Since $-t_n \to 0$ and $t_n \to 0$, part (a) gives $s_n \to 0$.

Tags: Standardε–N / ε–δ

Used in Theorem 9.7, Theorem 11.2

Exercise 8.6

Let $(s_n)$ be a sequence in $\mathbb{R}$.

1. Prove $\lim s_n = 0$ if and only if $\lim |s_n| = 0$.
2. Observe that if $s_n = (-1)^n$, then $\lim |s_n|$ exists, but $\lim s_n$ does not.

Hint

(a) $|s_n - 0| = \bigl||s_n| - 0\bigr|$, so the two definitions say exactly the same thing.

Solution

1. $|s_n - 0| = |s_n| = \bigl||s_n| - 0\bigr|$ for every $n$, so “$|s_n - 0| < \varepsilon$ for $n > N$” and “$\bigl||s_n| - 0\bigr| < \varepsilon$ for $n > N$” are the same statement. Hence $s_n \to 0$ iff $|s_n| \to 0$.
2. If $s_n = (-1)^n$ then $|s_n| = 1$ for all $n$, so $\lim|s_n| = 1$ exists, while $\lim(-1)^n$ does not exist (Example 4). (Part (a) is special to the limit $0$.)

Tags: Standardε–N / ε–δ

Used in Exercise 12.2

Exercise 8.7

Show the following sequences do not converge.

1. $\cos\left(\dfrac{n\pi}{3}\right)$
2. $s_n = (-1)^n n$
3. $\sin\left(\dfrac{n\pi}{3}\right)$

Hint

Imitate Example 4. (a) The values $1$ (at $n = 6k$) and $-1$ (at $n = 6k + 3$) recur forever, so take $\varepsilon = 1$. (b) A convergent sequence is bounded (Theorem 9.1), or argue directly. (c) $\frac{\sqrt3}{2}$ and $-\frac{\sqrt3}{2}$ recur forever; take $\varepsilon = \frac{\sqrt3}{2}$.

Solution

1. For $n = 6k$ the term is $\cos(2k\pi) = 1$ and for $n = 6k+3$ it is $\cos((2k+1)\pi) = -1$; both values occur for arbitrarily large $n$. Suppose the limit were $a$. Taking $\varepsilon = 1$, for all large $n$ we’d have $|s_n - a| < 1$, and choosing large $n$ with $s_n = 1$ and large $m$ with $s_m = -1$ gives $2 = |1 - (-1)| \le |1 - a| + |a + 1| < 2$, a contradiction.
2. $|s_n| = n$ is unbounded, but convergent sequences are bounded (Theorem 9.1). So $(s_n)$ does not converge.
3. For $n = 6k+1$ the term is $\sin(\frac\pi3 + 2k\pi) = \frac{\sqrt3}2$, and for $n = 6k+4$ it is $\sin(\frac{4\pi}3 + 2k\pi) = -\frac{\sqrt3}2$; both recur forever. If $s_n \to a$, take $\varepsilon = \frac{\sqrt3}2$: large $n$ and $m$ with these values give $\sqrt3 \le |s_n - a| + |a - s_m| < \sqrt3$, a contradiction.

Tags: Standardε–N / ε–δContradiction

Used in §7

Exercise 8.8

Prove the following (see Exercise 7.5):

1. $\lim\left[\sqrt{n^2+1} - n\right] = 0$
2. $\lim\left[\sqrt{n^2+n} - n\right] = \frac12$
3. $\lim\left[\sqrt{4n^2+n} - 2n\right] = \frac14$

Hint

(a) $\sqrt{n^2+1} - n = \frac{1}{\sqrt{n^2+1}+n} < \frac1{2n}$. (b) $\left|\frac{n}{\sqrt{n^2+n}+n} - \frac12\right| = \frac{\sqrt{n^2+n} - n}{2(\sqrt{n^2+n}+n)} < \frac{1}{8n}$, because $\sqrt{n^2+n} - n = \frac{n}{\sqrt{n^2+n}+n} < \frac12$ and the denominator exceeds $4n$. (c) is similar.

Solution

1. $0 < \sqrt{n^2+1} - n = \dfrac{1}{\sqrt{n^2+1} + n} < \dfrac1{2n}$, which is $< \varepsilon$ for $n > \frac1{2\varepsilon}$.
2. Write $\sqrt{n^2+n} - n = \dfrac{n}{\sqrt{n^2+n} + n}$. Since $n^2 + n < (n + \frac12)^2$, we have $\sqrt{n^2+n} - n < \frac12$. Then $\left|\sqrt{n^2+n} - n - \tfrac12\right| = \left|\frac{n}{\sqrt{n^2+n}+n} - \frac12\right| = \frac{\sqrt{n^2+n} - n}{2(\sqrt{n^2+n} + n)} < \frac{1/2}{4n} = \frac1{8n},$ because $\sqrt{n^2+n} + n > 2n$. This is $< \varepsilon$ for $n > \frac1{8\varepsilon}$.
3. Let $T = \sqrt{4n^2+n}$. Then $T - 2n = \dfrac{n}{T + 2n}$ and $T > 2n$, so $T - 2n < \frac n{4n} = \frac14$. Also $\left|T - 2n - \tfrac14\right| = \left|\frac{n}{T+2n} - \frac14\right| = \frac{T - 2n}{4(T + 2n)} < \frac{1/4}{16n} = \frac1{64n},$ which is $< \varepsilon$ for $n > \frac1{64\varepsilon}$.

Tags: Standardε–N / ε–δComputation

Exercise 8.9

Let $(s_n)$ be a sequence that converges. (This exercise is referred to in several places.)

1. Show that if $s_n \ge a$ for all but finitely many $n$, then $\lim s_n \ge a$.
2. Show that if $s_n \le b$ for all but finitely many $n$, then $\lim s_n \le b$.
3. Conclude that if all but finitely many $s_n$ belong to $[a, b]$, then $\lim s_n$ belongs to $[a, b]$.

Hint

(a) Suppose $s = \lim s_n < a$. Take $\varepsilon = a - s$. Then for large $n$, $s_n < s + \varepsilon = a$, which contradicts the hypothesis for all large $n$.

Solution

Let $s = \lim s_n$.

1. Suppose $s < a$ and let $\varepsilon = a - s > 0$. For all large $n$, $s_n < s + \varepsilon = a$, so $s_n < a$ for infinitely many $n$ (all large ones), contradicting that $s_n \ge a$ for all but finitely many $n$. Hence $s \ge a$.
2. Suppose $s > b$ and let $\varepsilon = s - b$. For all large $n$, $s_n > s - \varepsilon = b$, contradicting $s_n \le b$ for all but finitely many $n$. Hence $s \le b$.
3. If all but finitely many $s_n$ lie in $[a,b]$, then $s_n \ge a$ and $s_n \le b$ for all but finitely many $n$, so $a \le s \le b$ by (a) and (b).

Tags: Standardε–N / ε–δContradiction

Used in §9, Exercise 9.4, Exercise 9.9, §10, Exercise 11.9, Theorem 18.1

Exercise 8.10

Let $(s_n)$ be a convergent sequence with $\lim s_n > a$. Prove there is a number $N$ such that $n > N$ implies $s_n > a$.

Hint

Use $\varepsilon = \lim s_n - a$.

Solution

Let $s = \lim s_n > a$ and $\varepsilon = s - a > 0$. There is $N$ such that $n > N$ implies $|s_n - s| < \varepsilon$, so $s_n > s - \varepsilon = a$.

Tags: Standardε–N / ε–δ

Used in Theorem 9.9, Exercise 9.11