need a smart math person (3rd time asking); True or False quesion?

Question

True OR false?
Lim k-->∞ [1/(k(k+2))] < lim k-->∞[5/(k^2)]?

This can be rewritten as:
Lim k-->∞ [1/(k^2+2k))] < lim k-->∞[5/(k^2)]?

Answer 1

false, both limits are zero
the first one does go to zero faster than the second one, but they are both eventually zero

Answer 2

Calculate them explicitly. Both are equal to 0. Hence they're equal.

WHY are they both 0, you ask? Well, they're both uniformly positive and bounded above by 1/k. And 1/k goes to 0 because you can make it forever be less than epsilon just by waiting until k gets to be > 1/epsilon.

Answer 3

TRUE, it is less than because the k^2 absolutely swamps the 2k in the denominator as k ---> infinity and it becomes 1/k^2 < 5/k^2
Obviously both go to zero but the first term gets there faster

Answer 4

This is FALSE! Both limits are 0.
To see this try plugging larger and larger
values of k into each one. Both get smaller
and smaller as k gets arbitrarily large.
So both can be made as small as you like
if k is large enough.

Answer 5

FALSE: Even if it is coming from god instead of these calculator-dependent "smart math people".
Both limits are zero and hence equal.

Answer 6

Lim k-->∞ [1/(k(k+2))] < lim k-->∞[5/(k^2)] - true

Lim k-->∞ [1/(k^2+2k))] < lim k-->∞[5/(k^2)] - true

Answer 7

if you evaluate, the limits can not be determined exactly.
but i guess this is true, cuz on the left side you get a ->0/->0 form, on the right side you get a 5/->0. and then 5/->0 is obv greater than ->0/->0

Answer 8

True

Answer 9

true
denominator on the left is greater than on the right
value on the left is less than that of the right

Answer 10

IT is true.Just have a look at the graphs of these two equations.

http://akipic.com/uploads/6767e04fcd.jpg