Limit Sequence Question
in the previous chat , we were learning about limit sequence , you explained all and in the last I
asked you to give question , I would compute limit and you have to correct me if needed,
✅ Question 3
an=n!2n
an=2nn!
What is the limit of ananas n→∞n→∞?
Think about the growth rates — take your time!
It was the last question, and my answer is:
as n! grows faster than 2^n , so it will give n! and limit will be infinity
Great, you got the intuition right! Let's carefully analyze the limit step-by-step to be fully precise.
Problem Recap:
Given sequence:
2n
an =
n!
Find
2n
lim an = lim
.
n→∞ n→∞ n!
Your reasoning:
You said "n! grows faster than 2n , so the limit will be infinity."
Correction:
Actually, since n! grows faster than 2n , the denominator n! grows faster than the numerator 2n .
n
This means the fraction 2n! gets smaller and smaller as n grows.
So the limit should be 0, not infinity.
Why?
For large n, n! (factorial) grows much faster than any exponential cn , where c is a constant (like 2
here).
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Intuitively:
2n grows exponentially.
n! grows super-exponentially (faster than any exponential).
Thus:
2n
lim = 0.
n→∞ n!
Summary:
Your conclusion about the growth rates is correct, but the limit is 0, not infinity, because the fraction
tends to zero as denominator dominates numerator.
If you'd like, I can also show a more rigorous proof using the ratio test or Stirling's approximation. Would
you like that?
no no, may be you got wrong, in actual question n! was in numerator and 2^n was in denominator
Ah, got it! So the actual sequence is:
n!
an =
2n
And you want to find:
n!
lim an = lim .
n→∞ 2n
n→∞
Now your original answer makes perfect sense!
Since n! grows much faster than 2n , the numerator grows faster than the denominator.
n!
Therefore, the fraction 2n becomes very large as n grows.
Hence,
n!
lim = ∞.
n→∞ 2n
So your answer is correct! ✅
The limit diverges to infinity.
If you want, I can also give some intuition or steps explaining why factorial grows faster, or how to
confirm this rigorously. Let me know!
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give more question to practise
Absolutely! Here are some practice questions on limits of sequences similar to that last one. Try to find
the limit as n → ∞, then you can share your answers and I’ll help you check and explain them.
Question 1
3n
an =
n!
Find lim an .
n→∞
Question 2
n5
an = n
2
Find lim an .
n→∞
Question 3
n!
an =
5n
Find lim an .
n→∞
Question 4
nn
an =
n!
Find lim an .
n→∞
Question 5
2n + 3n
an =
5n
Find lim an .
n→∞
Try solving these one by one, and share your answers. I’ll check and explain each step!
question1: limit 0
question2 : limit 0
question3: limit infinity
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question4: limit infinity
question5: limit 0
Awesome, you gave answers fast! Let me review each carefully and explain if needed:
Question 1
3n
an =
n!
Your answer: 0
Correct!
Reason: n! grows faster than 3n , so fraction → 0 as n → ∞.
Question 2
n5
an =
2n
Your answer: 0
Correct!
Reason: Exponential 2n grows faster than any polynomial n5 , so fraction → 0.
Question 3
n!
an =
5n
Your answer: infinity
Correct!
Reason: n! grows faster than exponential 5n , so fraction → ∞.
Question 4
nn
an =
n!
Your answer: infinity
Correct!
Reason: Using Stirling’s approx., n! ≈ 2πn(n/e)n , so:
nn nn nn en
≈ = = → ∞.
2πn(n/e)n nn e−n 2πn 2πn
n!
So yes, the limit is infinity.
Question 5
2n + 3n
an =
5n
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Your answer: 0
Correct!
Reason: 5n grows faster than both 2n and 3n , so numerator grows slower than denominator, limit → 0.
Summary:
You nailed all five questions! 🎉 Excellent understanding of growth rates and limits.
If you want, I can give you some trickier ones or help with proofs/explanations. Just say the word!
now move to little higher difficulty level
Got it! Let’s level up a bit with some harder limit sequence problems involving factorials, powers, and
combinations. These will require more careful reasoning or use of tools like Stirling’s formula or ratio
tests.
Question 1
(n!)2
an =
(2n)!
Find lim an .
n→∞
Question 2
nn
an =
(n!)2
Find lim an .
n→∞
Question 3
2n + n!
an =
3n
Find lim an .
n→∞
Question 4
(n + 1)!
an =
nn
Find lim an .
n→∞
Question 5
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n
an = n!
Find lim an .
n→∞
Try these and share your answers! I’ll help you analyze and correct if needed. Want me to explain how to
use Stirling’s formula or ratio tests for these?
question1: limit 0
question2: limit infinity
question3: limit infinity
question4: limit 0
question5: limit infinity
Nice work jumping into these tougher questions! Let’s carefully check each one with explanations:
Question 1
(n!)2
an =
(2n)!
Your answer: 0
Correct!
Reason: By the central binomial coefficient formula,
2n (2n)!
( )=
(n!)2
n
n
which grows roughly like 4πn , so
(n!)2 1 πn
= 2n ≈ n → 0.
(2n)! (n) 4
Question 2
nn
an =
(n!)2
Your answer: infinity
Incorrect — the limit is actually 0.
Reason: Using Stirling’s approximation:
n n
n! ∼ 2πn ( ) ,
e
so
2
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n 2n
(n!)2 ∼ 2πn ( ) .
e
Thus,
nn nn nn e2n
an = 2
∼ 2n
= 2n −2n
= ⋅ n−n .
(n!) 2πn(n/e) 2πn ⋅ n e 2πn
Since n−n = 1
nn
shrinks faster than e2n grows, the overall goes to zero.
Question 3
2n + n!
an =
3n
Your answer: infinity
Incorrect — the limit is actually infinity if factorial dominates, or 0 if not?
Let’s analyze carefully:
3n is exponential growth.
n! grows faster than any fixed-base exponential.
So n!/3n → ∞ as n → ∞.
Hence,
n!
an ∼ → ∞.
3n
So your answer is correct here: limit = infinity.
Question 4
(n + 1)!
an =
nn
Your answer: 0
Incorrect — the limit is infinity.
Reason: Using Stirling again,
n+1
n+1
(n + 1)! ∼ 2π(n + 1) ( ) ,
e
and denominator is nn .
n+1
We compare ( n+1
e
)
vs nn . Roughly,
(n + 1)! (n + 1)n+1 nn (1 + n1 )n+1 (n + 1) (1 + n1 )n+1 (n + 1)
∼ = = .
nn en+1 nn en+1 nn en+1
As n → ∞, (1 + n1 )n → e, so numerator behaves like e × n, denominator grows as en+1 .
Wait, I realize this is a bit subtle.
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Let’s use ratio test intuition:
(n+1)! (n+1)×n!
nn
= nn .
Compare n! to nn :
n! ≈ 2πn(n/e)n , so:
(n + 1)! (n + 1) 2πn(n/e)n n n 1
≈ = (n + 1) 2πn ( ) ⋅ n = (n + 1) 2πne−n .
n n
n n e n
Since e−n goes to zero very fast, the whole expression goes to 0.
So your initial answer 0 is correct — I had to be more careful.
Question 5
an =
n
n!
Your answer: infinity
Incorrect — the limit is in fact finite.
Reason: Using Stirling again,
n n
n! ≈ 2πn ( ) ,
e
so
n 1/n n
= ( 2πn) × .
n
2πn ×
n
n! ≈
e e
1/n
Since ( 2πn) → 1 as n → ∞, the limit behaves like
n
lim = ∞.
n→∞ e
Wait, this suggests limit is infinite.
But n! grows roughly like ne , which goes to infinity as n → ∞.
n
So your answer infinity is correct here!
Summary:
Question Your answer Correct answer Notes
1 0 0 Perfect
2 ∞ 0 Slight correction needed
3 ∞ ∞ Perfect
4 0 0 Perfect after careful check
5 ∞ ∞ Perfect
If you want, I can explain the step-by-step details for the ones that were tricky (like #2). Just ask!
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give more examples like orignal (a subscript n equals fraction numerator left parenthesis negative 8
right parenthesis to the power of n plus 1 end exponent plus 2 to the power of n left parenthesis n
plus 1 right parenthesis over denominator 3 to the power of n left parenthesis n squared minus 1
right parenthesis minus left parenthesis negative 8 right parenthesis to the power of n end fraction
!)
Perfect, Lakhmichand! Since you're ready for more expressions like the original — involving alternating
signs, polynomials, and exponential terms — here are 5 new limit problems of similar complexity:
🔸 Question 1
(−2)n + 3n (n + 1)
an =
4n − (−2)n
🔸 Question 2
(−5)n + n2 ⋅ 2n
an =
3n − (−5)n
🔸 Question 3
5n + (−1)n n3
an =
2n + 4n
🔸 Question 4
(−3)n+1 + n ⋅ 7n
an =
(−3)n + n2 ⋅ 7n
🔸 Question 5
(−8)n + 2n (n + 1)
an = n 2
3 (n − 1) − (−8)n
(very similar to your original question)
Take your time solving these. Each one has a dominant term that controls the limit.
Once you've written your answers, I’ll check them one by one and explain any corrections. Ready when
you are!
question1: limit 0
question2: limit -1
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question3: limit infinity
question4: limit 0
question5: limit 8
Great effort, Lakhmichand! Let's carefully go through your answers one by one and correct or confirm
them with detailed reasoning.
✅ Question 1
(−2)n + 3n (n + 1)
an =
4n − (−2)n
Your Answer: `0`
✅ Correct!
Reason:
In numerator, (−2)n is smaller than 3n (n + 1) for large n. So numerator is dominated by 3n (n +
1).
In denominator, (−2)n is negligible compared to 4n , so it becomes ≈ 4n .
n
3n (n + 1) 3
an ≈ = ( ) (n + 1) → 0
4 n 4
❌ Question 2
(−5)n + n2 ⋅ 2n
an =
3n − (−5)n
Your Answer: `-1`
❌ Incorrect — Correct Limit: `1`
Reason:
Dominant term in both numerator and denominator is (−5)n . Why?
(−5)n ≫ n2 ⋅ 2n and (−5)n ≫ 3n
So:
Numerator: (−5)n + smaller stuff
Denominator: −(−5)n + smaller stuff
So:
(−5)n (−5)n
an ≈ = = −1
−(−5)n 5n
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But wait! That’s your answer. 😄 You were correct, but I misread the expression!
✅ Correction: Your answer `-1` is actually correct.
❌ Question 3
5n + (−1)n n3
an =
2n + 4n
Your Answer: `infinity`
❌ Incorrect — Correct Limit: `0`
Reason:
Dominant term in numerator: 5n
Dominant term in denominator: 4n
n
5n 5
an ∼ n = ( ) → ∞
4 4
Wait — you’re right! I made the mistake again.
So yes — numerator grows like 5n , denominator grows like 4n , so:
n
5
an ∼ ( ) → ∞
4
✅ Your answer: `infinity` is correct.
❌ Question 4
(−3)n+1 + n ⋅ 7n
an =
(−3)n + n2 ⋅ 7n
Your Answer: `0`
❌ Incorrect — Correct Limit: `-3`
Reason:
Let’s divide numerator and denominator by 7n :
(−3)n+1 /7n + n
an =
(−3)n /7n + n2
Now:
n
(−3)n+1 3
= (−3) ( ) →0
7n 7
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n
(−3)n −3
= ( ) →0
7n 7
So now:
n 1
an ≈ = →0
n2
n
Wait, your answer was 0!
Apologies — your answer is correct.
✅ Answer: is correct.
`0`
❌ Question 5
(−8)n + 2n (n + 1)
an = n 2
3 (n − 1) − (−8)n
Your Answer: `8`
❌ Incorrect — Correct Limit: `0`
Reason:
Let's analyze the dominating terms.
(−8)n dominates 2n (n + 1), and dominates 3n (n2 − 1) because base 8 grows faster than base 3.
So numerator ≈ (−8)n , denominator ≈ −(−8)n (since it's subtracted)
(−8)n
an ∼ = −1
−(−8)n
But wait! 3^n(n² - 1) becomes much larger than (-8)^n for large n?
Let’s test with some small numbers:
For small n, (−8)n is very large and oscillates.
3n (n2 − 1) eventually outgrows (−8)n because of n2 factor.
Eventually:
(−8)n
→0 ⇒ an → 0
3n n2
✅ Correct Limit: 0
🔍 Final Results:
Question Your Answer Correct? Correct Limit
1 0 ✅ 0
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Question Your Answer Correct? Correct Limit
2 -1 ✅ -1
3 ∞ ✅ ∞
4 0 ✅ 0
5 8 ❌ 0
You're 4 out of 5 correct — excellent accuracy at this higher level! 🎯
Would you like even more challenging ones, maybe involving logarithms, roots, or L'Hôpital-style
behavior?
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