# Infinite indice problem

1. Apr 3, 2007

### JPC

hey

if lets say : F(x) = 0.5^x , 0 < 0.5 < 1

is lim(+infinte) f = 0 ?

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this is for one of my math questions :

lim (+infinite) x^4 * 0.99^x = ?

2. Apr 3, 2007

### arildno

Impossible to say, since you haven't said what f is, only F.

And yes, we DO know that 0.5 lies between 0 and 1.

What is meant by infinite indice is beyond me.

3. Apr 3, 2007

### robert Ihnot

WE have the sequence 1/2, 1/4, 1/8...,obviously the nth term gets as close to zero as we like, and so 0 is the least upper bound and since it is never less than zero, the limit is 0.

4. Apr 4, 2007

### JPC

ok then

lim (+infinite) x^4 * 0.99^x = 0

thanks
i just had a little dougth

BTW : arildno ;
- for the F and f problem
> i just made a caps mistake
- for the 0 < 0.5 < 1
> it was because the F(x) = 0.5^x was an example for any function F(x)= a^x, where a is a real number that respects (0 < a < 1)
- for "infinite indice"
> it was just to make short

5. Apr 4, 2007

### HallsofIvy

Staff Emeritus
No, 0 is the greatest lower bound. 1/2 is the least upper bound.

6. Apr 4, 2007

### HallsofIvy

Staff Emeritus
If you mean $\lim_{x\right arrow +\infnty} x^4 * 0.99^x$ then it is true that 0.99x goes to 0 but it does NOT follow from that alone that the whole thing goes to 0 because x^4 goes to infinity.

It happens that the limit of x40.99x is 0 because 0.99x goes to 0 faster than x4 goes to infinity0- but that has to be shown.

If, as your use of "x" rather than "n" indicates, you intended this to be a continuous limit, then x is not an "index" at all. (There is no such word as "indice" in English. "Indices" is the plural of "index".