Question

Give an example of a function f(x) defined on the closed interval [1,5] so that f(3) = 1 and lim x->3 f(x) does not exist but lim x->(3) {f(x)}^2 does exist and equals 4. Explain your example.

Answer 1

looks like you need a piecewise function

Answer 2

but i am a bit confused. and also since \(1^2=1\) and not 4, it am not sure how \(f(3)^2=4\) is possible

Answer 3

Thank you for your reply. Yes I understand I need a piecewise function, but f(x) can only be either 2 or -2. Am I correct?

Answer 4

yeah this is trickier than i thought at first, or maybe not possible. if \(f(3)=1\) i cannot figure out how \(f^2(3)=4\)

Answer 5

Yeah, but thanks satellite! I guess I'll head over and talk to my professor about it.

Answer 6

by definition \(f^2(x)=(f(x))^2\) and so it seems that \(f^2(3)=(f(3))^2=1^2=1\)
for the limit to exist the value of the function at a number \(a\) must be the same as the value of the limit of the function as x approaches \(a\) so in your case it must be true that \(f^2(3)=4\)

Answer 7

if you get an answer please post, because it is a mystery to me

Answer 8

Thanks, I will keep you posted.

Answer 9

Hi Satellite, this is how the function should look like! Correct me if I'm wrong.

Answer 10

oh silly me, i thought it had to be continuous at \(x=3\)
yes of course this is correct

Answer 11

in fact you could have lots of different example, like
\[ f(x) = \left\{
\begin{array}{lr}
x-1 & : 1\leq x\leq 3\\
1-x & : 3<x\leq 5
\end{array}
\right.\]

Answer 12

no, that wouldn't work

Answer 13

\[f(x) = \left\{
\begin{array}{lr}
x-1 & : 1<3\\
1 & : x=3\\

1-x & : 3<x\leq 5
\end{array}
\right.\]

Answer 14

Yes, there are in fact infinitely many examples for this question. Thanks for the insightful share!