limit of infinite sequence

not sure to put this in pre calculus or calculus, sorry if mistaken

Find the following limit.

Given that where ,

find

the answer given is:

since

since

therefore, when

therefore

i don't understand how and is found.

what i find instead is and

using that, i also find

but please help explain the steps, as the steps in the answers are pretty different from those that i use, im afraid that i've mistaken

Re: limit of infinite sequence

Quote:

Originally Posted by

**muddywaters** the answer given is:

since

as

goes from

to

,

goes from 0 to -1, so the absolute value goes from 0 to 1.

since

as

goes from

to

,

goes from 0 to -1 and back to zero, but the point where it is -1,

, is excluded so the absolute value is between 0 to 1.

Since and have absolute value strictly less than 1, their n-th powers go to zero as n goes to infinity. So the numerator goes to zero and the denominator goes to 0-3 = -3, and therefore the limit is zero.

- Hollywood

Re: limit of infinite sequence

"as goes from to , goes from 0 to -1, so the absolute value goes from 0 to 1"

that's how the answer says, but how can we say that 'so the absolute value goes from 0 to 1'? that would be , which is the same as , but what we have found so far is actually only , not as well.

and the question gives , so taking that into account, , and that's how i get for the part of the question.

Re: limit of infinite sequence

All of the things you say are true. But when we say that , we're not saying that takes on all possible values that fit the inequality. We're only saying that whatever values of takes on, the inequality holds.

So for every value of , , it is true that , even though, as you said, there are other values of which would also make the statement true.

- Hollywood