Quote:
Originally Posted by nein12
I am having a few issues with these annoying questions. If anyone have time to figure these questions out, I would really appreciate it.
I need lim x> 0 for this piecewise function
f(x)
1/ (x3)^2 if x not equal to 3
2 if x = 3
I can get f(3), but I don't know how to get the limit. 1/(x3)^2 becomes undefined when x = 3. Is that mean the limit doesn't exist, or did I screw thing up?
I don't know why f(x) as x > 2 does not exist...
f(x)
sqaure root (4x^2) if 2 < x < 2
4x  sin(3.14x) if 2 equal or less than x
Quick concept reminder...
What's g o f (2) when g(2) is 2 and f(2) is also 3? Do I use 3 and put it into g(3) or do I add them up and get 1 as answer?
lim x > 1+ sqaure root (1x) I thought it's zero but graph says otherwise. Help!
Again, thanks in advance. I don't usually bother people like this, but this is kinda urgent.

Your first question:
lim x> 0 for the piecewise function f(x) where
f(x) = 1/ (x3)^2 if x not equal to 3
= 2 if x = 3
Since you are taking the limit as x goes to 0, f(x) is defined as 1/(x3)^2 at that point, so simply evaluate f(0) = 1/(03)^2 = 1/9
I do not understand your next question.
"lim x > 1+ sqaure root (1x) I thought it's zero but graph says otherwise. "
lim as x goes to 1 (from the positive side) of sqrt(1  x). sqrt(1  x) is not defined under the real numbers for x > 1.
Hope these help some.