Limits to the infinity 2

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Limits to the infinity 2

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\[\lim_{x \rightarrow \infty} {{2x^2 - 3} \over {4x^3 + 5x}}\]
Divide the numerator and denominator by x^3, and send to infinity.
Factor an \(x^3\) top and bottom and cancel. Then go to the limit.

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Other answers:

so, the answer is 0? coz the numerator > denominator?
0
The numerator will go to 0, and the denominator will go to 4.
ops, sorry invert this numerator < denominator
no. infinity is not a number. you cannot say that infinity^2 is greater than infinite by itself
0/4 = 0
ah, ok!
\[\lim_{x \rightarrow \infty}\frac{2/x-3/x^2}{4+5/x^2}=\frac{\lim_{x \rightarrow \infty}2/x+\lim_{x \rightarrow \infty}-3/x^2}{\lim_{x \rightarrow \infty}4+\lim_{x \rightarrow \infty}5/x^2}=\frac{0-0}{4+0}\]
Right! Thanks!
np

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