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 [itex]r(t)=\left< \frac { sint }{ t } ,\frac { { e }^{ 2t }-1 }{ t } ,{ t }^{ 2 }ln(t) \right>[/itex]
Re-define [itex]r(t)[/itex] to make it right continuous at [itex]t=0[/itex]
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