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Old 06-18-2009   #2
MAS1

 
Join Date: Dec 2008
Posts: 249
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Quote:
Originally Posted by nein12 View Post
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/ (x-3)^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/(x-3)^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 (4-x^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 (1-x) 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/ (x-3)^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/(x-3)^2 at that point, so simply evaluate f(0) = 1/(0-3)^2 = 1/9

I do not understand your next question.

"lim x -> 1+ sqaure root (1-x) 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.
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