Question

Limits to the infinity 2

Answer 1

\[\lim_{x \rightarrow \infty} {{2x^2 - 3} \over {4x^3 + 5x}}\]

Answer 2

Divide the numerator and denominator by x^3, and send to infinity.

Answer 3

Factor an \(x^3\) top and bottom and cancel. Then go to the limit.

Answer 4

so, the answer is 0? coz the numerator > denominator?

Answer 5

0

Answer 6

The numerator will go to 0, and the denominator will go to 4.

Answer 7

ops, sorry
invert this numerator < denominator

Answer 8

no. infinity is not a number. you cannot say that infinity^2 is greater than infinite by itself

Answer 9

0/4 = 0

Answer 10

ah, ok!

Answer 11

\[\lim_{x \rightarrow \infty}\frac{2/x-3/x^2}{4+5/x^2}=\frac{\lim_{x \rightarrow \infty}2/x+\lim_{x \rightarrow \infty}-3/x^2}{\lim_{x \rightarrow \infty}4+\lim_{x \rightarrow \infty}5/x^2}=\frac{0-0}{4+0}\]

Answer 12

Right! Thanks!

Answer 13

np