# Thread: Graph this piecewise function and use it to determine the values of a for which the l

1. ## Graph this piecewise function and use it to determine the values of a for which the l

$\displaystyle \begin{displaymath} f(x) = \left\{ \begin{array}{lr} 2-x & if x<-1\\ x & if -1<=x<1\\ (x-1)^2 & if x>=e1 \end{array} \right. \end{displaymath}$
I'm supposed to graph this piecewise function and use it to determine the values of a for which the limit of f(x) exists as x approaches a.

I'm not entirely sure how to start this question. Looking on the graph I'm thinking the only place where the limit of f(x) as x approaches a doesn't exist is -1 and 1. Could someone please explain?
That is the graph. So does the limit of f(x) as x approaches a exist everywhere except 1 and -1 ?

2. Originally Posted by iamanoobatmath
$\displaystyle \begin{displaymath} f(x) = \left\{ \begin{array}{lr} 2-x & if x<-1\\ x & if -1<=x<1\\ (x-1)^2 & if x>=e1 \end{array} \right. \end{displaymath}$
I'm supposed to graph this piecewise function and use it to determine the values of a for which the limit of f(x) exists as x approaches a.

I'm not entirely sure how to start this question. Looking on the graph I'm thinking the only place where the limit of f(x) as x approaches a doesn't exist is -1 and 1. Could someone please explain?
That is the graph. So does f(x) exist everywhere 1 and -1
Well, f(x) is defined over all the reals, but $\displaystyle \displaystyle\lim_{x\to a}f(x)$ does not exist at a=-1 and a=1 as you said. There's really not a lot to it. (If you want some more detail: at those points left and right limits exist, but they are not equal.) Nice presentation with LaTeX and graph.

3. Originally Posted by undefined
Nice presentation with LaTeX and graph.
Wolframlpha is a life safer when the bloody textbook and the solution manual only gives answers to odd questions.

Nice name, btw.