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Prove the limit of (4n² +9)/(3n² + 7n+11)

User Tijko
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Final answer:

The limit of the sequence (4n² +9)/(3n² + 7n+11) is proven to be 4/3 by dividing both the numerator and denominator by n² and taking the limit as n approaches infinity.

Step-by-step explanation:

The student is asking to prove the limit of the sequence (4n² +9)/(3n² + 7n+11). To do this, we leverage the fact that as n approaches infinity, terms that grow slower than n² become negligible in the fraction. Thus, we focus on the highest-degree terms in both the numerator and denominator.

To find the limit, divide both the numerator and the denominator by n²:

(4n² +9)/(3n² + 7n+11) = (4 + 9/n²)/(3 + 7/n + 11/n²).

Now take the limit as n approaches infinity. The terms featuring n in the denominator will approach zero, leaving us with:

Limit [n → ∞] of (4 + 9/n²)/(3 + 7/n + 11/n²) = 4/3.

This shows that the limit of the given sequence is indeed 4/3 as n becomes very large.

User John Moutafis
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