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

Mathematics

1 answer:

hram777 [196]2 years ago

4 0

Answer:

$\lim_{n\rightarrow +\infty } \frac{4n^{3}-5}{9n^{3}+7}=\frac{4}{9}$

Step-by-step explanation:

$\lim_{n\rightarrow +\infty } \frac{4n^{3}-5}{9n^{3}+7}$

Pull out n³

$=\lim_{n\rightarrow +\infty } \frac{n^{3}\times \left( 4-\frac{5}{n^3} \right) }{n^{3}\times \left( 9+\frac{7}{n^3} \right) }$

Simplify by n³

$=\lim_{n\rightarrow +\infty } \frac{ \left( 4-\frac{5}{n^3} \right) }{ \left( 9+\frac{7}{n^3} \right) }$

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

Then

$\lim_{n\rightarrow +\infty } \frac{ \left( 4-\frac{5}{n^3} \right) }{ \left( 9+\frac{7}{n^3} \right) }=\frac{4-0}{9+0}$

$=\frac{4}{9}$

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