A function including a infinite Limit

[Beny.Maleky]

If we consider the function f, defined like this :
f(x) = Lim(x/n) , which (n->infinity)
Then we know for all small 'x's (in comparison to numbers approaching to infinity), f(x) is equal to 0, but if we calculate the Lim of this function when x approaches to infinity, f(x) would be 1, which it means: L = Lim(Lim (x/n) = 1, which (n -> infinity , x-> infinity)
Now if we graph the introduced function, it will show us that all the f(x) are zero, but we know somewhere in the x-axis, f(x) will approach to 1 . For example lets suppose that this happens in x=a.
so by saying that, I want to know that how can we determine a place for x=a on x-axis, which f(x)=1?
I mean where does that change happen? How does that change happen? Is y=f(x) continuous?

 

[Romsek]

[MATH]\forall x \in \mathbb{R}, \lim \limits_{n \to \infty} \dfrac{x}{n} = 0[/MATH]
You're either taking the limit to infinity or you're not. You can't have it both ways.

 

[Beny.Maleky]

Thank you! I understood the answer.

 

[Steven G]

x is a real number. and it is not tending to infinity. However, n is approaching infinity. So x/n approaches 0 as n approaches infinity.

 

[Steven G]

Is y = f(x) continuous? Not if it is not defined like when n=0.

 

[Beny.Maleky]

x is a real number. and it is not tending to infinity. However, n is approaching infinity. So x/n approaches 0 as n approaches infinity.

I understand what you are saying, but what if both x and n tend to infinity? Isn't it possible?

 

[topsquark]

I understand what you are saying, but what if both x and n tend to infinity? Isn't it possible?

You used the notation f(x). This says that x is a finite value, not a limit tending to infinity. You can define a new value [math]f = \lim_{x, n \to \infty} \dfrac{x}{n}[/math] but that isn't a function of x.

-Dan