$(\varepsilon, \delta)$ proof for $f(x)=\begin{cases} \frac{3-x}{2} & x<1 \\ x & x \geqslant1 \end{cases} $ 
Show using $(\varepsilon, \delta)$- definition of continuity that 
  $$f(x)=\begin{cases} 
      \frac{3-x}{2} & x<1 \\
      x &  x \geqslant1 
   \end{cases}
$$
  is continuous at $x=1$.

I had a problem like this on my class and wasn't aware of the $(\varepsilon, \delta)-$definition for continuity and approached the problem a bit differently.
In order for $f(x)$ to be continuous at $x=1$ we would have to have the left and right-handed limits equal to each other.
Since we have:
$\lim_{x\to1^-} \frac{3-x}{2}=1$ (1)
$\lim_{x\to1^+} x= 1$ (2)
we can proceed on proving this using $(\varepsilon, \delta)$ for the limits.
for (1) we can pick $\delta=2\varepsilon$ and since $|\frac{3-x}{2}-1| = |\frac{-x+1}{2}| =|\frac{x-1}{2}| \overset{\mathrm{(x > 1)}}{=}
 \frac{x-1}{2} < \frac{\delta}{2} = \frac{2\varepsilon}{2} =\varepsilon$ the limit holds.
similarly, for (2) we can pick $\delta=\varepsilon$ and since $|x-1|< \delta=\epsilon$ the limit holds as well.
I know this isn't what they asked for, but shouldn't it be pretty much the same thing?
 A: I would say that you are almost done with the problem, but not quite there yet.  What you have is $\delta_\text{left}$ and $\delta_\text{right}$ such that the following holds:  if $0<x-1<\delta_\text{right}$ then $|f(x)-f(1)|<\varepsilon$, and if $0<1-x<\delta_\text{left}$ then $|f(x)-f(1)|<\varepsilon$.  All that's left is finding some $\delta$ such that $|f(x)-f(1)|<\varepsilon$ whenever $|x-1|<\delta$.
A: 1) $x\ge 1$: 
$|f(x)-f(1)|=|x-1|$;
2) $x<1:$
$|f(x)-f(1)|= |\frac{3-x}{2} -1| =$
$|\frac{1-x}{2}| \lt |x-1|;$
3) Let $\epsilon >0$ be given;
Choose $\delta = \epsilon$;
4) For $x \ge 1$:
$|x-1| < \delta$ implies 
$|f(x)-f(1)| =|x-1|< \delta=\epsilon$;
For $x < 1:$
$|x-1| <\delta$ implies
$|f(x)-f(1)| < |x-1| < \delta= \epsilon$.
Altogether:
For $\delta=\epsilon$:
$|x-1|\lt \delta$ implies $|f(x)-f(1)| <\epsilon$.
A: !OOOPS!
Actually on rereading you forgot the most important part!  You must actually show that $f(1) = \lim_{x\to 1} f(x)$.  You show the left and right limits are equal to each other but you didn't show they are both equal to $f(1)$.  (Which is easy.... just point out $f(1) = 1$ by definition. and you've already shown $ \lim_{x\to 1^-} f(x) = \lim_{x\to 1^+} f(x) = 1$.)
====== but read on for my original post ======
It's the same thing.  Exactly.  
The definition of "$f$ is continuous at $x=a$" is word for word "for any $\epsilon > 0$ there is a $\delta > 0$ so that if $0< |x-a|< \delta$ then $|f(x)-f(a)|< \epsilon$".
The definition of $\lim_{x\to a}f(x)= L$ is word for word "for any $\epsilon > 0$ there is a $\delta > 0$ so there if $0< |x-a| < \delta$ then $|f(x) - L | < \epsilon$".
And so the definition for $\lim_{x\to a}f(x) = f(a)$ is wor for word "for any $\epsilon > 0$ there is a $\delta > 0$ so that if $0< |x-a|< \delta$ then $|f(x)-f(a)|< \epsilon$" which is precisely with utterly no change or variation the definition for "$f$ is continuous at $x=a$".
