Limit of a Sequence

Context: FIT1058_MOC · the long-run value a sequence settles to · defined with nested quantifiers ($\forall \epsilon \exists N\forall n$) · underlies infinite series and growth

Quick Revision

• 🎯 Objective: the value terms get arbitrarily close to and stay close to ➔ $\forall \epsilon \exists N\forall n>N:|a_n-\ell |<\epsilon$.
• 📦 Core Components: $\epsilon$ tolerance ➔ $N$ cutoff (depends on $\epsilon$) ➔ all later terms within $\epsilon$.
• ⚡ Key Constraint: quantifier order ($\forall \epsilon \exists N$) is essential; “eventually”, not “always”.

📝 Core

1. The Definition

• Limit ➔ ${\lim }_{n\to \infty }{a}_n=\ell \iff \forall \epsilon >0\exists N\forall n>N:|a_n-\ell |<\epsilon$.
• Read ➔ for any target distance $\epsilon$, some point $N$ beyond which all terms lie within $\epsilon$.

2. Order & Scope

• $N$ after $\epsilon$ ➔ $N$ may depend on $\epsilon$ (smaller $\epsilon$ → larger $N$).
• Eventually ➔ only terms beyond $N$ matter; finitely many early strays are irrelevant.
• $\epsilon >0$ ➔ arbitrarily small, never 0.

3. Non-Convergence

• Unbounded ➔ $(n),(n^2)$ grow without bound.
• Oscillating ➔ $((-1{)}^n)$ bounded, no limit.
• Split ➔ odd/even terms to different values.

Key identities:

$\text{Given }\epsilon >0,\text{ choose }N>\frac{1}{\epsilon }.n>N\Rightarrow \frac{1}{n}<\epsilon \Rightarrow \left|\frac{1}{n}-0\right|<\epsilon .$

When It Flips: an infinite Geometric Series is the limit of partial sums (exists iff $|r|<1$); for diverging sequences, growth order replaces a finite limit. Only the infinite tail matters, never a finite prefix.

📊 Exam Execution Trace

Applied Exercise

Problem: Prove ${\lim }_{n\to \infty }\frac{1}{n}=0$ and find $N$ for $\epsilon =0.01$. Derivation Proof / Hand-Calculation Walkthrough:

$$\begin{array}{rl}\text{given }\epsilon >0,N& >\frac{1}{\epsilon }\Rightarrow n>N\Rightarrow \frac{1}{n}<\epsilon \\ \epsilon =0.01& :N=101\Rightarrow a_{102}=\frac{1}{102}<0.01✓\end{array}$$

Final Extracted Output: $N=\lceil 1/\epsilon \rceil$ works for every $\epsilon$ ⟹ limit 0.

⚠️ Common Mistakes

• 💡 Order matters ➔ $\forall \epsilon \exists N$ lets $N$ depend on $\epsilon$; swapping to $\exists N\forall \epsilon$ demands one cutoff for all $\epsilon$ — a far stronger, usually false claim.

🧠 Active Recall

State the $\epsilon$–$N$ definition and prove $\lim \frac{1}{n}=0$.

• Hint: $N$ as a function of $\epsilon$.

Answer

• Short answer: $\forall \epsilon >0\exists N\forall n>N:|a_n-\ell |<\epsilon$; take $N>\frac{1}{\epsilon }$ for $\frac{1}{n}$.
• Why: Exists for every $\epsilon$ ➔ so the limit is 0.

Why must $N$ depend on $\epsilon$, and why only terms "beyond $N$"?

• Hint: Quantifier order + eventually.

Answer

• Short answer: $\forall \epsilon \exists N$ picks $N$ after $\epsilon$; tighter $\epsilon$ needs later $N$.
• Why: Tail only ➔ finitely many early strays can’t affect the limit.