What is the limit of (2/3)^n as x approaches infinity?

[gillgill]

1) lim (2/3)^n
x-> infintiy
2) lim (4/3)^n
x-> infinity

 

[A_I_]

1) the answer is 0
because 2/3 = 0.6666 < 1 ---> lim=0

2) the answer is 1
because 4/3 = 1.3333
and when this number tends to infinity it is considered equal to one thus the answer is 1

 

[A_I_]

i forgot to tell you also that u made a mistake n should tend to infinity and not x. :)
joe

 

[dextercioby]

The first limit is +\infty...

Daniel.

 

[chroot]

because 4/3 = 1.3333
and when this number tends to infinity it is considered equal to one thus the answer is 1

Uhhh... what?

Absolutely false. The answer to this limit is infinity.

If you multiply a quantity by a fraction less than one an infinite number of times, it tends to zero.

If you multiply a quantity by a number greater than one an infinite number of times, it tends to infinity.

In no way, whatsoever, is 4/3 ever "considered equal to one."

- Warren

 

[Jameson]

\lim_{x\rightarrow\infty} (\frac{4}{3})^x = \infty

If you can't see that, graph the equation. Or if you don't have a calulator, use the first derivative for critical points, and you find none.

The first answer is correct, but the second one is infinity.


Jameson

 

[A_I_]

yeah, i doubted it first.. i thought it was wrong. You corrected it.
thanks guys :)

 

[dextercioby]

It can be elegantly proven using the famous commutation rule:

\ln \lim_{x}f(x) =\lim_{x}\ln f(x)

Daniel.

 

[ToxicBug]

The first limit is +\infty...

Daniel.

No its not,
1) lim (2/3)^n
x-> infintiy
=2^n/3^n

so its zero.

 

[dextercioby]

I was referring to post #2...

Daniel.

 

[Icebreaker]

\ln \lim_{x}f(x) =\lim_{x}\ln f(x)

That's odd, I have never heard of this (maybe I did, but have forgotten). Time to dig up the ol' calculus book.

 

[dextercioby]

It comes about because the natural logarithm is an strictly ascending function on its domain.


Daniel.

 

[Hurkyl]

You're forgetting the most important part -- continuity.