Squeeze theorem for sequences

What it actually says

The bounds only need to hold eventually (for $n \ge N_0$ ). Limits ignore any finite chunk at the start.
The two outer sequences must converge to the same limit $L$ . If they go to different limits, the theorem says nothing.

WHY it's true — derivation from scratch

Setup. Fix any $\varepsilon > 0$ . We must find $N$ so that $|b_n - L| < \varepsilon$ for all $n \ge N$ .

Step 1 — Use $a_n \to L$ . Why? This is the lower wall. By definition of the limit, there is $N_1$ with $|a_n - L| < \varepsilon \quad (n \ge N_1) \;\Longrightarrow\; L - \varepsilon < a_n.$ We only keep the left half — that's all the lower wall can give us.

Step 2 — Use $c_n \to L$ . Why? This is the upper wall. There is $N_2$ with $|c_n - L| < \varepsilon \quad (n \ge N_2) \;\Longrightarrow\; c_n < L + \varepsilon.$

Step 3 — Combine with the squeeze. Why? Now we stack the inequalities. Let $N = \max(N_0, N_1, N_2)$ so all three facts hold at once. For $n \ge N$ : $L - \varepsilon < a_n \le b_n \le c_n < L + \varepsilon.$ Reading the outermost pieces: $L - \varepsilon < b_n < L + \varepsilon \;\Longleftrightarrow\; |b_n - L| < \varepsilon.$

Step 4 — Conclude. Why? We produced the required $N$ for an arbitrary $\varepsilon$ . That is exactly the definition of $\lim_{n\to\infty} b_n = L$ . $\blacksquare$

Worked examples

Worked example 3)

$n$ th root: $b_n = \sqrt[n]{n}$ — uses squeeze on $\ln$

Step — take logs. Why? Roots are hard; logs turn them into a quotient: $\ln b_n = \frac{\ln n}{n}$ .
Step — bound $\ln n$ . Why? For $n\ge1$ , use the standard inequality $\ln n \le 2(\sqrt n -1)$ , so $0 \le \frac{\ln n}{n} \le \frac{2(\sqrt n -1)}{n} \le \frac{2}{\sqrt n}$ .
Step — squeeze. $0 \le \frac{\ln n}{n} \le \frac{2}{\sqrt n} \to 0$ , so $\ln b_n \to 0$ , hence $b_n \to e^0 = 1$ .
Conclusion: $\displaystyle \lim_{n\to\infty}\sqrt[n]{n} = 1.$

Common mistakes (steel-manned)

Active recall

Recall Quick self-test (cover the answers)

State the three conditions needed to conclude $b_n\to L$ by squeeze. → $a_n\le b_n\le c_n$ eventually; $a_n\to L$ ; $c_n\to L$ (same $L$ ).
Why is one-sided bounding not enough? → It can't stop the sequence escaping on the other side.
What limit does $\frac{\cos(n^2)}{n}$ have, and which walls? → $0$ , walls $\pm\frac1n$ .
True/false: bounds must hold from $n=1$ . → False, only for $n\ge N_0$ .

Squeeze theorem (sequences) — statement

a_n\le b_n\le c_n

for all

n\ge N_0

and

a_n\to L

c_n\to L

, then

b_n\to L

Key requirement on the two outer sequences

They must converge to the SAME limit

L

Why must the bound only hold "eventually"?

Limits ignore any finite number of initial terms.

\lim_{n\to\infty}\frac{\sin n}{n}

and the walls used

0

; walls

-\frac1n \le \frac{\sin n}{n}\le \frac1n

\lim_{n\to\infty}\frac{n!}{n^n}

0

(bounded above by

\frac1n

, below by

0

\lim_{n\to\infty}\sqrt[n]{n}

1

(take logs:

\frac{\ln n}{n}\to0

Why does one-sided bounding fail to give a limit?

The sequence could escape unboundedly on the unbounded side.

\varepsilon

-step that lower wall

a_n

supplies

L-\varepsilon < a_n

for

n\ge N_1

\varepsilon

-step that upper wall

c_n

supplies

c_n < L+\varepsilon

for

n\ge N_2

Index used to combine all three facts

N=\max(N_0,N_1,N_2)

Recall Feynman: explain it to a 12-year-old

Imagine you're in a lift squeezed between your tall mom and your tall dad, and they both walk into the same room. You can't go to a different room — you're stuck between them, so you end up in the same room. A sequence trapped between two others that both head to the number $L$ must also land on $L$ . We use this when the middle thing (like $\sin n$ ) wiggles too much to follow directly — we just put neat walls around it that both calm down to the same spot.

Connections

Limit of a sequence (epsilon-N definition) — the squeeze proof is pure $\varepsilon$ – $N$ .
Bounded sequences — supplies the walls ( $|\sin n|\le 1$ , etc.).
Algebra of limits (sum, product, quotient) — alternative tool when the sequence is tame.
Monotone convergence theorem — another existence-of-limit tool.
Squeeze theorem for functions — the continuous analogue; same $\varepsilon$ logic.
Standard limits ( n-th roots, n!/n^n, ln n / n ) — classic squeeze applications.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Squeeze theorem ka idea ekdam simple hai: agar koi sequence $b_n$ do doosri sequences $a_n$ aur $c_n$ ke beech phasi hui hai, aur dono outer sequences ek hi number $L$ par ja rahi hain, to beech wali $b_n$ ko bhi majboori mein $L$ par hi jana padega. Socho teen log lift mein — niche wala aur upar wala dono ek hi floor par rukte hain, to beech wala bhi wahin ruk jayega. Bas yahi "sandwich" logic hai.

Yeh kaam aata hai jab middle sequence bahut messy ho, jaise $\frac{\sin n}{n}$ . $\sin n$ to hamesha $-1$ se $1$ ke beech naachta rehta hai, uska direct limit lena tough hai. Lekin hum likh dete hain $-\frac1n \le \frac{\sin n}{n} \le \frac1n$ . Dono walls $0$ par ja rahe hain, isliye beech wala bhi $0$ par — done!

Sabse important condition yaad rakhna: dono walls ka limit SAME hona chahiye. Agar ek wall $0$ par ja raha hai aur doosra $5$ par, to theorem kuch nahi batata — $b_n$ kahin bhi ho sakta hai. Aur ek baat: inequality sirf "eventually" (yaani kisi $N_0$ ke baad) honi chahiye, shuru ke kuch terms se farak nahi padta, kyunki limit ko early terms ki parwah nahi hoti. Last tip — jab inequality ko divide ya multiply karo, sign ka dhyan rakho; positive ( $n>0$ ) se divide karoge to inequality ki direction safe rehti hai.

What it actually says

WHY it's true — derivation from scratch

Worked examples

Common mistakes (steel-manned)

Active recall

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Calculus III — Sequences & Series

Connections