# Is this function 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.

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#### 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.

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#### shmoe

Homework Helper
Benny said:
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?

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#### NateTG

Homework Helper
steven187 said:
does anybody know how to show something does not approach something else on latex?
$$\lim_{a \rightarrow b} f(a) \neq c$$

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

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#### mathwonk

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

#### Benny

Thanks for the help guys.

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