1. ## Limits and continuity

any help with this question would be muchly appreciated

2. ## Re: Limits and continuity

How rigorous does this need to be? For instance, are you suppose to use the epsilon-delta definitions? It is very easy to apply in this case:

For all $\epsilon >0$, choose $\delta = \frac{1}{2}$ so that $0<|x-3|<\frac{1}{2} \implies x\not \in \mathbb{Z} \implies f(x)=2 \implies |f(x)-2|=0 < \epsilon$. So the limit is $2$.

You can easily generalise this to arbitrary $a\in \mathbb{Z}$ instead of $3$. And by generalise I mean replace every instance of $3$ with $a$ and you'll get a valid proof.

The limit for b) is also two. You need minor modification to the proof above. For example, take $\delta$ to be the distance for $\pi$ (or $a\not \in \mathbb{Z}$) to the nearest integer.

In light of parts a) and b), it should be obvious if and where the function is continuous.

3. ## Re: Limits and continuity

For x any number "close" to 3, but not equal to 3, x is NOT an integer so f(x)= 2. Therefore $\lim_{x\to 3}f(x)= ?$

For x any number "close" to $\pi$, but not equal to $\pi$, x is NOT an integer so f(x)= 2. Therefore $\lim_{x\to 3}f(x)= ?$