Squeeze theorem for sequences: Question bank — Hinglish deep dive

4.3.2 · D5 · Maths › Calculus III — Sequences & Series › Sequences ke liye Squeeze theorem

Spot the error

sin n \to 0

because

\frac{s i n n}{n} \to 0

." — find the flaw.

Flaw yeh hai ki peeche divide kar diya:

\frac{s i n n}{n} \to 0

\frac{1}{n}

factor ke collapse hone se aata hai,

sin n

se nahi; actually

sin n

ka koi limit nahi hai, yeh

[- 1, 1]

mein oscillate karta rehta hai.

0 \leq \frac{n !}{n ^{n}} \leq \frac{1}{n}

, and since the upper wall

\frac{1}{n} \to 0

I'm done — no need for the lower wall." — is the reasoning complete?

Yahan yeh kaam karta hai sirf isliye kyunki explicit lower wall

0

bhi stated hai aur woh bhi

\to 0

; bina lower wall ke sirf upper bound se kuch prove nahi hota, toh "

0 \leq

" zaroori hai, decorative nahi.

"To bound

\frac{s i n n}{n}

, multiply

- 1 \leq sin n \leq 1

- \frac{1}{n}

to get

- \frac{1}{n} \leq \frac{- s i n n}{n} \leq \frac{1}{n}

." — spot the sign slip.

Negative number

- \frac{1}{n}

se inequality multiply karne par dono signs flip hone chahiye, result hoga

\frac{1}{n} \geq \frac{- s i n n}{n} \geq - \frac{1}{n}

; flip karna bhoolna classic error hai, isliye sirf positive quantities jaise

n > 0

se divide/multiply karo.

"Walls are

a_{n} = - \frac{1}{n} \to 0

and

c_{n} = 2 - \frac{1}{n} \to 2

; since

b_{n}

is trapped and both walls converge,

b_{n}

converges." — where is the mistake?

Dono walls converge karti hain, haan, lekin alag-alag limits (

0

aur

2

) par; squeeze ko ek common limit chahiye, toh yeh conclusion unjustified hai.

"The bound

a_{n} \leq b_{n} \leq c_{n}

fails at

n = 1, 2, 3

, so the squeeze theorem cannot be used." — correct or not?

Sahi nahi hai — theorem ko bound sirf

n \geq N_{0}

ke liye chahiye; finitely many early indices par failure irrelevant hai, bas

N_{0} = 4

lo.

"Since

\frac{l n n}{n} \to 0

, I can conclude

n n \to 0

." — trace the error.

Log ka limit

0

hai, aur

n n = e^{l n n / n} \to e^{0} = 1

0

nahi; log ko undo karna (exponentiate karna) bhool jaana answer ko galat jagah le jaata hai.

Why questions

Why must both outer sequences converge to the same limit?

Kyunki

ε

-band argument ke liye ek single target

L

chahiye: lower wall deti hai

L - ε < a_{n}

aur upper deti hai

c_{n} < L + ε

usi same

L

ke around; do alag limits do alag bands deti hain jo

b_{n}

ko pin nahi kar paatein.

Why is it enough for the inequality to hold "eventually" (for

n \geq N_{0}

Limit sirf sequence ki tail se decide hoti hai; finitely many terms ko change ya drop karna convergence ya limit value ko change nahi karta, toh early violations harmless hain.

Why do we bound

sin n

instead of trying to compute

lim sin n

directly?

Kyunki

sin n

ka koi limit nahi hai — yeh

[- 1, 1]

mein hamesha oscillate karta rehta hai; squeeze us messy oscillation ko sidestep karne deta hai aur

\pm \frac{1}{n}

jaisi tame walls mein wrap karta hai jinhein hum evaluate kar sakte hain.

Why does dividing

- 1 \leq sin n \leq 1

n

preserve the direction of the inequalities?

Kyunki

n > 0

hai, aur positive number se divide karne par order maintain rehta hai; sirf negative number se divide karne par

\leq

signs reverse hote hain.

Why can't a single upper bound ever prove a limit exists?

Ek upper bound sirf yeh batata hai ki sequence kitni high ja sakti hai; yeh nahi batata ki yeh kitni low dip karti hai, toh sequence upper wall ke neeche rehte hue bhi neeche diverge ho sakti hai.

Why is

∣ b_{n} - L ∣ < ε

equivalent to

L - ε < b_{n} < L + ε

Absolute value distance measure karta hai, toh "

b_{n}

L

ε

ke andar hai" ka literal matlab hai ki

b_{n}

open interval

(L - ε, L + ε)

mein hai — dono statements ek hi fact ke do tariqe hain.

Why does the proof set

N = max (N_{0}, N_{1}, N_{2})

rather than just one of them?

Humein teenon facts (bound, aur har wall ka apne band ke andar hona) simultaneously sach chahiye; sabse bada index lena guarantee karta hai ki inme se har ek already "switch on" ho chuki hai.

Edge cases

b_{n} = L

is constant and

a_{n} \leq L \leq c_{n}

with both walls

\to L

, does the squeeze still apply?

Haan — constant sequence ek bilkul valid

b_{n}

hai; yeh trapped hai aur trivially

L

par converge karti hai, theorem ke consistent hai (chahe theorem ki zaroorat na bhi ho).

What if

a_{n} = c_{n}

for all

n

(the walls coincide)?

Tab

a_{n} \leq b_{n} \leq a_{n}

force karta hai ki

b_{n} = a_{n}

exactly ho, toh

b_{n}

seedha wall ka limit le leta hai — squeeze ek equality ban jaata hai jisme hilne ki koi jagah nahi.

Can the walls oscillate (not be monotone) and the squeeze still work?

Haan — theorem monotone walls kabhi require nahi karta, sirf yeh chahiye ki har wall

L

par converge kare; e.g.

a_{n} = - \frac{1 + ( - 1 ) ^{n}}{n}

phir bhi

\to 0

hai aur valid wall hai.

What happens if

b_{n}

touches or equals a wall infinitely often?

Kuch break nahi hota — inequalities non-strict (

\leq

) hain, toh

b_{n}

chahe kitni bhi baar

a_{n}

c_{n}

ke equal ho, phir bhi

L

par squeeze hoga.

If both walls diverge to

+ \infty

(

a_{n}, c_{n} \to \infty

) with

a_{n} \leq b_{n} \leq c_{n}

, what can we say about

b_{n}

Hum conclude kar sakte hain ki

b_{n} \to + \infty

bhi — yeh "squeeze to infinity" variant hai, lower wall

a_{n} \to \infty

b_{n}

ko oopar dhakele se driven hai; finite-limit squeeze theorem directly apply nahi hota.

Does the squeeze theorem require

b_{n}

to be defined by a formula?

Nahi —

b_{n}

koi bhi real sequence ho sakti hai (chahe uski koi closed form na ho); sirf do-taraf ka bound aur common wall limit mayne rakhte hain.

If only

a_{n} \leq b_{n}

holds (no upper wall) but

a_{n} \to L

, can we salvage anything?

Sirf ek one-sided statement: hum paate hain

lim inf b_{n} \geq L

, matlab

b_{n}

ultimately

L

se neeche nahi ja sakta; yeh convergence se weaker hai aur complete hone ke liye ek upper wall chahiye.

Recall One-line survival kit

Do walls, same limit, eventually hold karna, $b_{n}$ ko dono taraf se trap karna — inme se koi ek miss karo aur theorem chup ho jaata hai. Do walls, same limit, eventual bound, two-sided — charon? ::: Tab aur sirf tab $b_{n} \to L$ guaranteed hai.

Connections

Squeeze theorem for sequences — parent statement aur proof jinhein ye traps stress-test karte hain.
Limit of a sequence (epsilon-N definition) — kyun "eventually" aur $ε$ -band reasoning kaam karta hai.
Bounded sequences — jahan se walls $\pm \frac{1}{n}$ aati hain.
Algebra of limits (sum, product, quotient) — tame alternative jab koi squeeze zaruri na ho.
Squeeze theorem for functions — upar ke har trap ka continuous cousin.
Standard limits ( n-th roots, n!/n^n, ln n / n ) — examples mein assume kiye gaye limits.

Question bank — Squeeze theorem for sequences

True or false — justify

Spot the error

Why questions

Edge cases

Connections