The Two-Phase Proof Strategy
Writing an \( \epsilon-N \) proof is like a magic trick. The audience only sees the smooth performance (Formal Proof), but the real work happens backstage (Scratch Work).
Interactive Proof Builder
Prove that \( \lim \frac{3n+1}{7n-4} = \frac{3}{7} \)
Phase 1: Scratch Work (Backstage)
Goal: Start with \( |s_n - L| \lt \epsilon \) and solve for \( n \).
Step 1: Simplify \( |s_n - L| \)
Step 2: Set up inequality \( \lt \epsilon \)
Solve for \( n \):
Phase 2: Formal Proof (On Stage)
Now write it forwards. Fill in the blanks.
Let \( \epsilon \gt 0 \).
Choose \( N = \) .
Then for any \( n \gt N \), we have \( n \gt \frac{19}{49\epsilon} + \frac{4}{7} \).
Rearranging this gives \( 7n - 4 \gt \) .
Which implies \( \frac{19}{7(7n-4)} \lt \epsilon \).
Therefore, \( |s_n - L| \lt \epsilon \). Q.E.D.
The Art of Estimation
Sometimes solving for \( n \) exactly is too hard (e.g., with \( n^3 \)). Instead, we find a simpler, smaller denominator to make the fraction bigger.
Example: To bound \( \frac{1}{n^3 - 6} \), we want \( n^3 - 6 \ge \text{something simple} \).
Blue: \( y = n^3 - 6 \) | Red: \( y = n^3/2 \). Notice Blue is above Red for \( n \gt 2 \).
Practice Problems
True/False
1. In an \( \epsilon-N \) proof, we choose \( \epsilon \) first, then find \( N \).
False. \( \epsilon \) is given (arbitrary), we must find \( N \) that works for any given \( \epsilon \gt 0 \).
2. The scratch work is where we work backward from the conclusion to find \( N \).
True. This is the standard strategy to discover the relationship between \( N \) and \( \epsilon \).
3. If we find an \( N \) that works for a specific \( \epsilon \) (e.g., \( \epsilon = 0.1 \)), we have proven the limit.
False. The proof must work for all \( \epsilon \gt 0 \), not just one specific value.
Fill in the Blank
1. To prove \( \lim s_n = s \), we need to show that for every \( \epsilon \gt 0 \), there exists a natural number \( N \) such that if \( n \gt N \), then \( |s_n - s| \lt \) ___.
Answer: \( \epsilon \)
2. When proving a limit for a rational function, we often need to find a(n) ___ bound for the numerator and a lower bound for the denominator.
Answer: Upper
3. The inequality \( |s_n - s| \lt \epsilon \) is equivalent to \( s - \epsilon \lt s_n \lt \) ___.
Answer: \( s + \epsilon \)
Full Problems
1. Prove \( \lim \frac{1}{n^2} = 0 \).
Proof: Let \( \epsilon \gt 0 \). Choose \( N = \frac{1}{\sqrt{\epsilon}} \).
If \( n \gt N \), then \( n \gt \frac{1}{\sqrt{\epsilon}} \implies n^2 \gt \frac{1}{\epsilon} \implies \frac{1}{n^2} \lt \epsilon \).
Thus \( |\frac{1}{n^2} - 0| = \frac{1}{n^2} \lt \epsilon \).
2. Prove \( \lim \frac{3n+2}{2n-1} = \frac{3}{2} \).
Scratch Work: \( |\frac{3n+2}{2n-1} - \frac{3}{2}| = |\frac{2(3n+2) - 3(2n-1)}{2(2n-1)}| = |\frac{6n+4 - 6n+3}{4n-2}| = \frac{7}{4n-2} \).
We want \( \frac{7}{4n-2} \lt \epsilon \iff 4n-2 \gt \frac{7}{\epsilon} \iff 4n \gt \frac{7}{\epsilon} + 2 \iff n \gt \frac{7}{4\epsilon} + \frac{1}{2} \).
Proof: Let \( \epsilon \gt 0 \). Choose \( N \gt \frac{7}{4\epsilon} + \frac{1}{2} \). Then for \( n \gt N \), the steps reverse to show \( |s_n - 3/2| \lt \epsilon \).
3. Prove \( \lim \frac{n+6}{n^2-6} = 0 \).
Scratch Work: For \( n \gt 3 \), \( n^2 - 6 \gt n^2/2 \) (since \( n^2 \gt 12 \)). Also \( n+6 \lt 2n \) for \( n \gt 6 \).
So \( |\frac{n+6}{n^2-6}| \lt \frac{2n}{n^2/2} = \frac{4}{n} \). We want \( \frac{4}{n} \lt \epsilon \implies n \gt \frac{4}{\epsilon} \).
Proof: Let \( \epsilon \gt 0 \). Choose \( N = \max(6, \frac{4}{\epsilon}) \). Then for \( n \gt N \), \( |\frac{n+6}{n^2-6}| \lt \frac{4}{n} \lt \epsilon \).
4. Prove \( \lim \frac{\sin n}{n} = 0 \).
Proof: We know \( -1 \le \sin n \le 1 \), so \( |\sin n| \le 1 \).
Thus \( |\frac{\sin n}{n} - 0| = \frac{|\sin n|}{n} \le \frac{1}{n} \).
Let \( \epsilon \gt 0 \). Choose \( N = \frac{1}{\epsilon} \). If \( n \gt N \), then \( \frac{1}{n} \lt \epsilon \), so \( |\frac{\sin n}{n}| \lt \epsilon \).