calculus question

Question

lim h(t), where h(t)= (cos(t)-1)/(t^2). note that h(t) is even, that is, h(t)=h(-t).

t--0

 

t=  (+-)0.002   (+-)0.0001   (+-)0.00005   (+-)0.00001

 

My answers..

h(t)= .4999           .5                   .5                .5

 

 

was I right?

Answer 1

If we try "plugging in" t = 0, we get 0/0, which is an indeterminate form for which L'Hopital's Rule can be applied.  

 

lim(t→0)[(1-cost)/(t²)] = lim(t→0)[sint/(2t)]

 

                                            = (1/2)lim(t→0)[(sint)/t]

 

                                            = (1/2)(1)

 

                                            = 1/2