Question

Hi everyone...I need a little help...could someone discuss with me why the limit as x goes to infinity of ((2^x)/(7^x))is zero?
I realize that it would appear that if you plopped down an infinity sign intot he numerator and denominator, that the denom would approach infinity faster essentially making the limit approach zero, but another way of looking at it is that infinity over infinity is improper and I would need to use L'hopital rule making it eventually 2/7...I think! Can anyone discuss this with me?

Answer 1

(2^x)/(7^x) = (2/7)^x

Answer 2

the limit as x --> inf for (2^x)/(7^x) is the same as

the limit as x---> inf for (2/7)^x

Answer 3

as x gets bigger and bigger, (2/7)^x gets smaller and smaller, so that explains why the limit is zero

Answer 4

Hi Jim...is it possible to explain it in another way? For some reason I can't see why raising a fraction to an infinity power would make the limit zero, unless of course it is as simple as the denomiator reaching infinity faster than the numerator...am I close?

Answer 5

yes the denominator gets bigger faster than the numerator

Answer 6

basically if you think of (2/7)^x as a graph and make up the table for it, you'll see that (2/7)^x is a decreasing function over its entire domain

Answer 7

it will start large, then get smaller and smaller as x ---> infinity

Answer 8

it will never be negative or zero, but it will get closer and closer to 0

Answer 9

ok...for some reason was thinking that it would make the fraction an infinity over infinity and then I would have to invoke ala Mr. Hospital...lol

Answer 10

you could do that, but you'd be left with


ln(2)*2^x
---------
ln(7)*7^x


after deriving both top and bottom, but you're not anywhere near simplifying that doing it that way

Answer 11

so instead, you would rewrite it into (2/7)^x, then use this rule

\[\Large \lim_{x\to\infty}\left(\frac{a}{b} \right )^x = 0, \ \text{where} \ a < b\]

and a, b are constants

Answer 12

ok...one thing that is still bother me then...if you can plop down an infinity for both the numerator and denominator and then say the bottom is growing faster, when would you ever reach a case of infinity over infinity and then be forced to use L'ophital?

Answer 13

I can't think of any specific examples at the moment, but there are a lot of times when the derivatives of a function are easier to work with compared to the original function (ex: x^3 --> 3x^2) and hopefully something cancels out to make things simpler

Answer 14

however, in the case of 2^x, the derivative of that has a 2^x in it, so it really doesn't go away after you derive it

Answer 15

Notice that for anything less than \(1\), if you square it, it gets smaller. \[
0.99\times0.99=0.9801
\]

Answer 16

\[\Large
\frac{2^x}{7^x} = \frac{e^{x\ln2}}{e^{x\ln7}} = e^{(\ln2-\ln7)x}
\]As \[
x\to \infty
\]Then \[
(\ln2-\ln7)x \to -\infty
\]

Answer 17

\[\Large
\lim_{x\to \infty}e^{-x}=0
\]

Answer 18

If you want to get even more formal, you'd use the definition of the limit.

Answer 19

so basically when d/dx of 2^x/7^x becomes 2x/7xafter lophital, you would still have infinity over infinity, so if you took the 2nd derivative by invoking lophital again, wouldn't d/dx of 2x/27 become 2/7? lol...am I just crazy? lol :o)

Answer 20

well you'd have ln(2) and ln(7) in there as well, but yes

Answer 21

You don't want to use l'Hospital on this.

Answer 22

2^x
----
7^x

would become


ln(2)*2^x
---------
ln(7)*7^x

after applying L'Hospitals rule

Answer 23

okay...so basically I made my derivative incorrectly...so I really just need to remember that if the top and bottom of a fraction are both going to infinity, if the bottom goes there faster, the limit is zero but if the top went faster, the limit would be infinity and the series or whatever would diverge right?

Answer 24

you got it

Answer 25

yay!! Thanks Jim! and the other guy too!

Answer 26

you're welcome