## anonymous 5 years ago Limits to the infinity 2

1. anonymous

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

2. anonymous

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

3. anonymous

Factor an $$x^3$$ top and bottom and cancel. Then go to the limit.

4. anonymous

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

5. anonymous

0

6. anonymous

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

7. anonymous

ops, sorry invert this numerator < denominator

8. anonymous

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

9. anonymous

0/4 = 0

10. anonymous

ah, ok!

11. anonymous

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

12. anonymous

Right! Thanks!

13. anonymous

np