# 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: