Question

Please help! Find the limit as x approaches infinity of (1+(6/x))^x

Answer 1

1.

easy to see; as x becomes larger and larger, notice what happens to 6/x. (ignore the exponent for the moment)

6/1 = 6.
6/2 = 3
6/6 = 1
6/12 = 1/2
6/100 = 3/50
6/600 = 1/100
6/1000 = 1/1000

basically, as 6/x where x -> infinity, the value of 6/x will approach zero.

so basically that means we have
(1+0)^x
well thats easy to simplify
(1)^x
and 1 to any exponent is just 1..
1^2 = 1
1^10000000000000 = 1
etc. so thus the answer is 1.

Answer 2

Your reasoning makes sense! But when I submit it, it says its wrong.. I don't know if there is another way of solving it..?

Answer 3

The section we are working on is dealing with series. I know the fxn converges, so there has to be a set number that it goes to, but it isn't one...

Answer 4

it's definitely not 1, wolfram says its e^6, i see why but don't quite understand the reasoning they give.

http://www.wolframalpha.com/input/?i=lim+x+-%3E+inf+%281%2B%286%2Fx%29%29^x+

if you want to see.

(1+1/x)^x = e, and 6/x = e^6 i believe, not immediately sure why though

Answer 5

Oh..they raised it to the e and then did ln(fxn)..im assuming.? Haha but thank you so much for your help!!! It was greatly appreciated!!! (((: