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28 votes
28 votes
Lim
n->infinity 4n^3-5/9n^3+7

User Creitve
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1 Answer

11 votes
11 votes

Answer:


\lim_(n\rightarrow +\infty ) (4n^(3)-5)/(9n^(3)+7)=(4)/(9)

Explanation:


\lim_(n\rightarrow +\infty ) (4n^(3)-5)/(9n^(3)+7)

Pull out n³


=\lim_(n\rightarrow +\infty ) (n^(3)* \left( 4-(5)/(n^3) \right) )/(n^(3)* \left( 9+(7)/(n^3) \right) )

Simplify by n³


=\lim_(n\rightarrow +\infty ) ( \left( 4-(5)/(n^3) \right) )/( \left( 9+(7)/(n^3) \right) )

Remark :
\lim_(n\rightarrow +\infty ) (5)/(n^3) \right) } = 0 \ \. \text{and} \ \ \lim_(n\rightarrow +\infty ) (7)/(n^3) \right) } = 0

Then


\lim_(n\rightarrow +\infty ) ( \left( 4-(5)/(n^3) \right) )/( \left( 9+(7)/(n^3) \right) )=(4-0)/(9+0)


=(4)/(9)

User Sampada
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