Thread: Questions involving continuity

I think it's by definition, for a function to be continuous
lim f(x) as x --> a = f(a)
which also means
lim f(x) as x --> a+ = f(a) = lim f(x) as x --> a-
where lim means "limit as", x --> a+ means keeping x > a but taking it towards a and x ---> a- means x approaches a from its lower side.
So...
at x = 1
lim (x+1) as x-->a+ = lim(x^2+bx+c) as x-->a- where a = 1
1 + 1 = 1 + b + c or b + c = 1
Do the same for x --> 3+ and 3-. Find b and c.

Well, I tried doing the problem..

1=1^2+b+c
3=9+3b+c

-6-3b=c
1=1+b+(-6-3b)
6=-2b
b=-3

Then I tried plugging b into the equation...

1=1-3+c
3=c

I got b right.. but c=4, how do I do it?

Nevermind, I figured out how to do the problem.. I'll keep on studying so hopefully I'll get the concepts. Thanks for the explanation, it helped a lot!