Question

find the trigonometric limits.

lim (1/t^2)sin^2(t/2) as t approaches to 0.

Answer 1

1/4

Answer 2

give me the procedure how u did

Answer 3

do you know how to calculate

\[\lim_{t\to0}\frac{\sin(t/2)}{t}\]

?

Answer 4

no.

Answer 5

can you do this ...

\[\lim_{t\to0}\frac{\sin(t)}{t}\]

Answer 6

isn't it 1?

Answer 7

yes

Answer 8

How about this then...
\[\lim_{t\to0}\frac{\sin(t/2)}{t}=\lim_{t\to0}\frac{1}{2}\frac{\sin(t/2)}{t/2}\]

Answer 9

why do you multiply 1/2 in front of sin(t/2)/t/2?

Answer 10

\[\lim_{t\to0}\frac{\sin(t/2)}{t}=\lim_{t\to0}\frac{\sin(t/2)}{2\cdot t/2}=\lim_{t\to0}\frac{1}{2}\frac{\sin(t/2)}{t/2}\]

Answer 11

thank you.

Answer 12

no problem