Question

the lim cosx/x^5=1
(x->-infinity)

Answer 1

is it cos(x/x^5) ?

Answer 2

must be anyway

Answer 3

no sorrry it's (cosx)

Answer 4

lim(cosx)/(x^5) = 1
x->infinity
!?

Answer 5

sorry one more time (cosx)(x^5 + 1) as x approaches -infinity

Answer 6

ok ok so its different than what you wrote

Answer 7

ooops yeah i meant plus one

Answer 8

(cosx)/(x^5 + 1) ?

Answer 9

is there division between them ?

Answer 10

mhm

Answer 11

yeah thats in there

Answer 12

-1 /(x^5 +1) <= cosx/(x^5 +1) <= 1/(x^5 + 1)
x-> -infinity
-1 / (-infinity ) <= cosx/(x^5 + 1) <= 1 / -infinity
0 <= cosx/(x^5 +1) <= 0

lim = 0

Answer 13

alright because the lim as cos goes to -infinity is -1?

Answer 14

no..
its the sandwich theorem (i think it is the name)
so you take the boundaries of cos
on the right side is -1 and left 1

Answer 15

okay gotch! that makes sense ! thank you so much for your help i get it now :)

Answer 16

you are welcome :)

Answer 17

wait one more question
don't you multiply with (x^5+1) it not divide

Answer 18

so i asked you if it is division or not you said it is

Answer 19

yeah it's (cosx)divided by (x^5+1)

Answer 20

so that is what i did ..

Answer 21

yeah but when you set it up to the inequality you odivided one by (x^5+1) aren't you supposed to multilpy since it's already in the bottom?

Answer 22

see the cos(x) has boundaries -1 , 1

now x^5 + 1 has no boundaries

so cos(x)/(x^5+1) might be as small as -1/(x^5 +1) or as large as 1/(x^5+1)

now if both -1/(x^5 +1) and 1/(x^5+1) approach the same number as x -> -infinity
it means that also cos(x)/(x^5 +1) approach this number which is the limit.

Answer 23

i seee :) thank you so much for your help i really appreciate it!

Answer 24

:)