anonymous
  • anonymous
Limits to the infinity 2
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\lim_{x \rightarrow \infty} {{2x^2 - 3} \over {4x^3 + 5x}}\]
anonymous
  • anonymous
Divide the numerator and denominator by x^3, and send to infinity.
anonymous
  • anonymous
Factor an \(x^3\) top and bottom and cancel. Then go to the limit.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.