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anonymous
 5 years ago
Limits to the infinity 2
anonymous
 5 years ago
Limits to the infinity 2

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow \infty} {{2x^2  3} \over {4x^3 + 5x}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Divide the numerator and denominator by x^3, and send to infinity.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Factor an \(x^3\) top and bottom and cancel. Then go to the limit.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so, the answer is 0? coz the numerator > denominator?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The numerator will go to 0, and the denominator will go to 4.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ops, sorry invert this numerator < denominator

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no. infinity is not a number. you cannot say that infinity^2 is greater than infinite by itself

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow \infty}\frac{2/x3/x^2}{4+5/x^2}=\frac{\lim_{x \rightarrow \infty}2/x+\lim_{x \rightarrow \infty}3/x^2}{\lim_{x \rightarrow \infty}4+\lim_{x \rightarrow \infty}5/x^2}=\frac{00}{4+0}\]
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