Question

lim t-->0 h(t), where h(t)= cos(t)-1/t^2.

for the table values t is +/- .002 , .0001 , .00005 , .00001

what do you guys get for the value of h(t) and limit??

when i plugged in .002 i got -.499999975

when i plugged in .0001 i got -.5
when i plugged in .00005 i got -.5
when i plugged in .00001 i got 0

so does the limit not exist since i got the value 0 for the last number??

Answer 1

Do you mean [cos(t)-1]/t^2 ?
or cos(t) - (1/t^2) ?

Is t in radians?

Only the first version has a finite limit.

If t is in radians, that limit is -1/2

You must have done the 0.00001 case incorrectly.

Answer 2

I assume the posted expression has missing parentheses, namely we're looking for:

lim t-->0 h(t),
where h(t)= ( cos(t)-1 )/t^2.

When dealing with limits, the calculator results are not always reliable, because there is only a limited number of digits in the working memory.

Your calculator displays to 9 significant digits, but incorrectly for the case of f(0.002), for which I get -0.499999833342

This means that for t=0.00001, it probably treats t² as zero, and the zero is a signal that something has gone wrong (like division by zero).

For t=0.00001, I get -0.50000004137019...

The limit is, in fact, -0.5. You can find it by using the d'Hôpital's rule:
Lim t→0 (cos(t)-1)/t²
=Lim (-sin(t))/(2t)
=Lim (-cos(t))/2
=-0.5

or by expanding cos(t) as a Taylor's series:
Lim t→0 (cos(t)-1)/t²
=Lim (1-t²/2+t4/4!-... -1)/t²
=Lim (-t²/2 + t4/24)/t²
=Lim (-(1/2) + t²/48 -...)
= -1/2