5-2t+sin(8t)/5t+cos(8t)

find the limit as t goes to negative infinity

Question

5-2t+sin(8t)/5t+cos(8t) 

find the limit as t goes to negative infinity

anonymous · Answer

is that
$$\frac{5-2t+\sin(8t)}{5t+\cos(8t)}$$

anonymous · Answer

ya

anonymous · Answer

don't mean to rush but i have 6 min to put in an answer on my online hw before it's considered late LOL procrastination at it's best

anonymous · Answer

-2/5

anonymous · Answer

OMG I LOVE YOU SO MUCH RIGHT NOW

zzr0ck3r · Answer

divide everything by t, you will see that after you run the limit, by the squeeze theorem that the trig functions over t will go to 0 and the constants will go to 0

anonymous · Answer

that was close ok now how did you come to that lol

zzr0ck3r · Answer

he went to wolfram alpha

anonymous · Answer

Yes :3

anonymous · Answer

so sin8t/t is 0

anonymous · Answer

same with cos8t/t

zzr0ck3r · Answer

$\frac{5-2t+sin(8t)}{5t+cos(8t)} =\frac{\frac{5}{t}-2+\frac{sin(8t)}{t}}{5+\frac{cos(8t)}{t}}$

zzr0ck3r · Answer

take the limit and every term with a t under it goes to 0

zzr0ck3r · Answer

you are left with -2/5

anonymous · Answer

is that because the denominator is going infinitly big?

zzr0ck3r · Answer

$-1\le \cos(ax) \le 1 \implies \frac{-1}{t}\le\frac{cos(ax)}{t}\le\frac{1}{t} $
taking the limit we see
$0\le\frac{cos(ax)}{t}\le0$ thus it must be 0