Is the Function f(x) = e^[x] Continuous?

[Benny]

Is this function continuous? Edit: Fixed - function should load now

$$f\left( x \right) = e^{\left[ x \right]}$$


Where the argument of the exponential is the greatest integer less than or equal to x.

For the function to be continuous at a point x = a we need $$\mathop {\lim }\limits_{x \to a} f(x) = f(a)$$. For this particular function, f(x) at x = a is just f(a) where a is an integer? But what about the limit? As far as I can see this function is like a sequence so that if I looked at the graph I would just see some dots. Is it possible to take any limits with this function? For example, can I actually take lim(x->3)f(x) and get a finite value? Further, could I take lim(x->2.5)f(x) for this particular function. Any help appreciated.

Edit: Fixed f(x)...it should look right now.

 

[Muzza]

Take a sequence x_n in (0, 1) with x_n -> 1. If f was continuous (at 1), then f(x_n) -> f(1) = e, but [x_n] = 0, i.e. f(x_n) = e^0 = 1 -> 1. Contradiction.

Surely a similar argument can show that f is discontinuous everywhere.

 

[shmoe]

As far as I can see this function is like a sequence so that if I looked at the graph I would just see some dots. Is it possible to take any limits with this function?

The graph isn't "dots" it's a step function. You know what the graph of g(x)=[x] looks like? Similar thing. For limits (and continuity) can you answer these questions about g(x)=[x]? (consider left and right handed limits seperately at integers)

 

[steven187]

let $$X\subseteq\Re$$, $$f: X\longrightarrow\Re$$ and $$x_{0}\in X$$ where X is the domain of the function, then f is continuous at $$x_{0}$$ if $$\forall x_{n}\in X$$,$$x_{n}\longrightarrow x_{0}$$ and $$f(x_{n})\longrightarrow f(x_{0})$$, you will see that if you choose a sequence on the interval [1,2] that converges to the intiger 2 for example $${x_{n}}={2-\frac{1}{n}$$ which gives $$f(x_{n})\longrightarrow f(2)$$, but a sequence on the interval [2,3] which converges to 2 for example $${x_{n}}={2+\frac{1}{n}$$ gives $$f(x_{n})\neq>f(2)$$

does anybody know how to show something does not approach something else on latex?

 

[NateTG]

does anybody know how to show something does not approach something else on latex?

$$\lim_{a \rightarrow b} f(a) \neq c$$

Hackish:
$$f(x)\not{\rightarrow}f(y)$$

 

[mathwonk]

a more interesting function would be e^(2pi)i[x]

 

[Benny]

Thanks for the help guys.