Question

How to calculate the limit of ininite functions?
e.g.
lim (n -> inifinite) (2^n + 1) / (3^n + 1)

Answer 1

for some large value of n, the +1 parts become meaningless and this reduces to:

2^N
---- = (2/3)^N
3^N

Answer 2

I know it becomes meaningless, the answer is 0 because theoretical you should divide the whole function with the largest denominator (3^n)

but it doesn't make sense to me.

when i try to do that trick on another sum i got stuck. Like

Lim (n to infinite) (2^n - 2^-n) / (2^n -1)

Answer 3

2^n - 2^-n
----------
2^n -1
^^ is that part of the exponent?

Answer 4

lim ((2^n / 2^n) - (2^-n / 2^n)
n-> -----------------
((2^n / 2^n) - (1 / 2^n))

becomes

lim 1 - (2^-n / 2^n)
n -> ---------------
1 - ( 1/ 2^n)

Answer 5

if we divide the top[ and bottom by 2^n

1 - 2^(-2n)
----------
1 - (1/2)^n

Answer 6

shall I rewrite the function with equotations so it becomes readable.. (didn't saw the button in the bottom of the screen)

Answer 7

readable is always helpful :) but if ascii text is easier for you just make sure that you do not do sloppy work :) include the ()s as needed is all

Answer 8

sorry.

Answer 9

so, your last answer looks good. but how to work from there?

Answer 10

someones at the door. be back in 5 k ? *sorry, for the inconvenience

Answer 11

negative exponents go to zero
vvv
1 - 2^(-2n)
----------
1 - (1/2)^n
^^^
exponents with a base <1 go to zero