Lim (2/3)^n x-> infintiy

1. Mar 13, 2005

gillgill

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

2. Mar 13, 2005

A_I_

because 2/3 = 0.6666 < 1 ---> lim=0

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

3. Mar 13, 2005

A_I_

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

4. Mar 13, 2005

dextercioby

The first limit is $$+\infty$$...

Daniel.

5. Mar 13, 2005

chroot

Staff Emeritus
Uhhh... what? :uhh:

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

6. Mar 13, 2005

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

7. Mar 13, 2005

A_I_

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

8. Mar 13, 2005

dextercioby

It can be elegantly proven using the famous commutation rule:

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

Daniel.

9. Mar 14, 2005

ToxicBug

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

so its zero.

10. Mar 14, 2005

dextercioby

I was referring to post #2...:tongue:

Daniel.

11. Mar 14, 2005

Icebreaker

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

12. Mar 14, 2005

dextercioby

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

Daniel.

13. Mar 14, 2005

Hurkyl

Staff Emeritus
You're forgetting the most important part -- continuity. :tongue2: