find the limit of (t^2 - 5t-9)/[2(t^4)+3(t^3)] if x approaches negative infinity.
in the final step after simplifying, why do you plug in a zero to get the answer 0/2=0? why not plug in infinity to get infinity/2=infinity?
the work for it can be seen in example 5 of this link : http://tutorial.math.lamar.edu/Classes/CalcI/LimitsAtInfinityI.aspx#Limit_LimAtInf_Ex1a
When solving a limit problem of this division form, generally divide the top and the bottom of the division by the highest power of t ( in this case thats t^4). This gives limit of [ (1/t^2) - (5/t^3) -(9/t^4) ]/ [ 2 + (3/t) ] Now inserting t to infinity. (1/t) goes to 0 leaving [ 0 - 0 - 0 ]/[2 + 0] = 0
why do you insert 0 when you can plug in infinity? sorry i'm confused...
You are not inserting 0. You are inserting infinity. Its just that 1/[infinity] = 0.
Do you understand why 1/infinity equals zero?
any number divided by infinity will appraoch 0 so you wont really have zero but something like 0.00000000000000000000000000000000005487 note "5478" is just a placer holder i just wanted to emphasize the rule
Join our real-time social learning platform and learn together with your friends!