The discussion revolves around the problem of redefining a vector function r(t) to ensure it is right-continuous at t=0. The function is composed of three components, each involving limits and potential indeterminate forms.
The conversation is active, with participants providing suggestions for the values of the components at t=0 and discussing the implications of indeterminate forms. There is acknowledgment of the need for continuity and the use of L'Hopital's Rule to evaluate limits.
Participants are navigating the definitions and limits of the component functions, particularly focusing on the behavior of the function as t approaches 0. The discussion includes the challenge of handling indeterminate forms in the context of the problem.
You need to define values for each of the three component functions so that r(0) exists, and ##\lim_{t \to 0^+} r(t)## exists and is equal to r(0).Sho Kano said:Homework Statement
Given r(t)=\left< \frac { sint }{ t } ,\frac { { e }^{ 2t }-1 }{ t } ,{ t }^{ 2 }ln(t) \right>
Re-define r(t) to make it right continuous at t=0
Homework Equations
The Attempt at a Solution
This is probably the simplest problem ever, but I don't even know what it's asking for. Right continuous as in right handed limit? How can I re-define it?
So something likeMark44 said:You need to define values for each of the three component functions so that r(0) exists, and ##\lim_{t \to 0^+} r(t)## exists and is equal to r(0).
Correct.Sho Kano said:So something like
x=1 when t=0
y=2 when t=0
z = 0 when t = 0 because 0*∞ is indeterminate if the 0 is not "constant"?SammyS said:Correct.
How about z when t = 0 ?
"Indeterminate" means you can't say what the value will be.Sho Kano said:z = 0 when t = 0 because 0*∞ is indeterminate if the 0 is not "constant"?
The limit is 0, by L'Hopital's Rule. So the way I'm re-defining it is making r(t) continuous for t is not 0, and make r(0)=<1,2,0>, like a piece-wise functionMark44 said:"Indeterminate" means you can't say what the value will be.
If you write ##t^2\ln(t)## as ##\frac{\ln(t)}{t^{-2}}##, you now have the indeterminate form ##[\frac{\infty}{\infty}]##, so you can use L'Hopital's Rule on it.
YesSho Kano said:The limit is 0, by L'Hopital's Rule. So the way I'm re-defining it is making r(t) continuous for t is not 0, and make r(0)=<1,2,0>, like a piece-wise function