How to make functions right-continuous

[Sho Kano]

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?

 

[Mark44]

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?

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]

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).

So something like
x=1 when t=0
y=2 when t=0

 

[SammyS]

So something like
x=1 when t=0
y=2 when t=0

Correct.

How about z when t = 0 ?

 

[Sho Kano]

Correct.

How about z when t = 0 ?

z = 0 when t = 0 because 0*∞ is indeterminate if the 0 is not "constant"?

 

[Mark44]

z = 0 when t = 0 because 0*∞ is indeterminate if the 0 is not "constant"?

"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.

 

[Sho Kano]

"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.

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

 

[Mark44]

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

Yes