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 06-17-2009 #1 nein12   Join Date: Jun 2009 Posts: 1 Calculus Questions... (I am stuck..) 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.
06-18-2009   #2
MAS1

Join Date: Dec 2008
Posts: 249

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/ (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.
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.

 06-20-2009 #3 hunter34 Guest   Posts: n/a Since this is a piecewise function you must evaluate $\lim_{x\rightarrow 3^+}f(x)$ and $\lim_{x\rightarrow 3^-}f(x)$ Both limits are equal to infinity. (Graph the equation, chances are you will learn why later.) For the second question: The way f(x) is defined as so: $ f(x)= \begin{cases} \sqrt{4-x^2}&-2 4x-\sin(3.14x)&2\le x \end{cases}$ So for x=2, you need to evaluate $4(2)-\sin(3.14\cdot2)$ If you want to find if the limit of f(x) as x approaches 2, you need to find the limit of f(x) as x approaches 2 from the positive side and the negative side. If these are equal, then the limit exists and is equal to that number. (Of course it is not so the limit does not exist). Third Question: $g\circ f(x)=g(f(x))$. From there, you should see what to do. Remember the definition of $1^+$. This means that you evaluate the function from the positive side of 1. Since this function is undefined for $x>1$ (Why?) the limit from the positive side does not exist and therefore, the limit of $\sqrt{1-x}$ at x=1 does not exist.
06-22-2009   #4
MAS1

Join Date: Dec 2008
Posts: 249

Quote:
 Originally Posted by hunter34 Since this is a piecewise function you must evaluate $\lim_{x\rightarrow 3^+}f(x)$ and $\lim_{x\rightarrow 3^-}f(x)$ Both limits are equal to infinity. (Graph the equation, chances are you will learn why later.) For the second question: The way f(x) is defined as so: $ f(x)= \begin{cases} \sqrt{4-x^2}&-2 4x-\sin(3.14x)&2\le x \end{cases}$ So for x=2, you need to evaluate $4(2)-\sin(3.14\cdot2)$ If you want to find if the limit of f(x) as x approaches 2, you need to find the limit of f(x) as x approaches 2 from the positive side and the negative side. If these are equal, then the limit exists and is equal to that number. (Of course it is not so the limit does not exist). Third Question: $g\circ f(x)=g(f(x))$. From there, you should see what to do. Remember the definition of $1^+$. This means that you evaluate the function from the positive side of 1. Since this function is undefined for $x>1$ (Why?) the limit from the positive side does not exist and therefore, the limit of $\sqrt{1-x}$ at x=1 does not exist.
For the first problem, nein12 asked for the limit as x goes to 0, not as x goes to 3 as you stated.

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