If r(t)=(sqrt(t+1)) i+ ((t^2−9)/(t+3)) j+ sin(−3πt) k.
Then find: lim r(t)=?

Question

If r(t)=(sqrt(t+1)) i+ ((t^2−9)/(t+3)) j+ sin(−3πt) k.
Then find: lim r(t)=?

anonymous · Answer

$$r(t)=\sqrt{t+1}  i+ \frac{ t ^{2}-9 }{ t+3 } j+\sin (-3\pi t) k$$

anonymous · Answer

Don't you need to specify the limiting condition? For example as t approaches a certain value.

anonymous · Answer

t to 1

anonymous · Answer

I'm rusty on my limit functions, but I'm not seeing anything that "blows up"here as t approaches 1. When t=1 sqrt(1+1) is defined. same with T^2-9 / t+3 . right up till the last part every thing is no problem. But sin ( -3 pi t) changes sign when approaching from one side versus the other. However, since it shrinks to zero from either side I think you're good here as well. So as t approaches 1 the limit of the vector is just the  same as plugging in t=1.

anonymous · Answer

So... Do you follow what I'm saying here?

anonymous · Answer

yes, thanks. i was a little confused as well, but you seemed to have clarified things up.