someone who's taking calculus.. can you help me?
yes i probably can-what is the question?
thanks! okay so in this link http://tutorial.math.lamar.edu/Classes/CalcI/LimitsAtInfinityI.aspx#Limit_LimAtInf_Ex1a look at example 4 and 5... why can't i use infinity to plug in for both examples? for some reason, in ex. 4 they plugged in infinity and in ex 5 plugged in a 0 to get the final answer 0...
very easy. you have different powers. when you have different powers after you will just plug in infinity you still will no be able to predics which one grows faster and which one slower. it just will no give you solution. infiniti in a power of two devided by infinity in the power of four does not make any sence
there are rules how to evaluate expression to deal with powers
but in ex 5, why couldnt i plug in the infinity even though there were different powers?
again-when powers are different you dont know what is bigger and what is smaller. therefore you dont know if it goes to 0 or to infinity
you take out the thing with the highest power. then you plug in infinity. after this step only.
wait in ex 4 they didnt plug in infinity. they plugged in 0 though...
is it because in ex4 the x approaches infinity and negative infininty that you plug in an infinity in the final step (to get - or +infinity/5) and in ex5 the x only approached negative infinity that they plugged in a 0 to get 0/2=0?
I dont know what are you talking about. both examples are identical.
* i meant in ex 5 they didnt plug in infinity
all you have to do is to remember-when there are different powers-you take out varible with the highest one and then plug in infinities
they sure did. you just missed this step
1 over infinity in a power of 2=0 etc
that is why they got 0 over 2 in the end
oh... okay... then in ex 4 why didn't the infinity be simplified to 0? sorry i'm so confused...
oh! is it because you cancel out all the rational numbers?
and in ex 4, once you cancel out all the rational numbers, then all you get is -x^3/5 so infinity^3 would still be infinity.... and in ex 5, once you cancel out al the rational numbers, there's nothing left in the numerator so it's 0?
they also plugged it in. 4 over infinity=0 plus infinity in the power of 3=infinity now we need to divide infinity by expression in the bottom. number over infinity-0, 0-5=-5. now we have to devide infinity by -5. we got -infinity
wow you are really very confused and mix everything up. lets start from the beggining. what points do you miss?
oh, so you understand what is going on there now?
wait.. i want to double check..
good. check and ask.
so first you simplify the equation with the usual dividing out biggest exponent ...
exactly. but you have to do it in one case-when you divide different powers by different powers. this is the only case
then, say you get [z^3 ((4/z)+z^3)]/[z^3((1/z^3)-5)] from the problem in ex 4. cancel out the z^3.
but the very idea is that then you get numbers divided by infinities which are obviously zeros
plug in infinity to the z's, they are essentially 0s..
so then you can deal with it simply
yes, sure. but do you understand why they are zeros?
because when you divide out the numbers by infinity, they become too small?
you dont. that is why you are so lost))) take calculator plz
now devide some number for very big number. lets say devide 5 by 10 000
now devide 5 by 100 000 and then by 1000000
see what is going on. every time you get number closer and closer to 0
so you need to get rid of powers, and bring what you can to numbers
then you can see it and manage it
did you get the idea?
i think so...
good. I am very lost with my exam so i will go back to it.
thanks for the help!
no prob i got some great help here too
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