lim (t-->infiniti) of [sin(-t)/t]?

B

bittersweete

New member
"the limit as t goes to infiniti of [sin(-t)/t]"

Help!
Well, I have the answer, its 0. But I don't know how to get to that answer. I know sin(t)/t=1 but that doesn't help much here.
Hmm...that makes sense that it would equal -1 if sin(-t) always = -sin(t) so the answer my prof gave must be wrong. Thanks!
 
M

Minny

New member
This makes no sense, because (-t/t) reduces to (-1), and since that's a constant, you would get: lim (t-->infinity) sin (-1) =-1.
 
W

wretrop

New member
Wait, no I got that confused with lim--->0.

0 is actually the answer. Sin(-t) only has limited values which is being divided by infinity.

It is considered then to be infinitesimally small, or, 0
 
B

BraveSirRobin

New member
No, your professor is right. The answer is 0. sin of anything will oscillate only between -1 and 1. Since the numerator is staying at a finite amount and the denominator is approaching infinity, the limit goes to 0. L'Hospital's rule only applies in cases of (0/0) or (infinity/infinity), which you don't have in this case, since sin is oscillating.

So, the answer is 0.
 
Top