"We can proceed as follows. The idea is to try to take "
"advantage of the Fundamental Trig Limit:" \quad lim_{ x rarr 0 } sin(x)/x \ = \ 1."
"Proceeding:"
\qquad lim_{ t rarr 0 } ( 1/t^2 ) sin^2( t/2 ) \ = \ lim_{ t rarr 0 } ( 1/t^2 ) [ sin( t/2 ) ]^2
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ = \ lim_{ t rarr 0 } [ 1/t cdot sin( t/2 ) ]^2
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ = \ lim_{ t rarr 0 } [ 1/2 cdot 2/t cdot sin( t/2 ) ]^2
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ = \ lim_{ t rarr 0 } [ 1/2 ]^2 cdot [ 2/t cdot sin( t/2 ) ]^2
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ = \ lim_{ t rarr 0 }1/4 cdot [ sin( t/2 ) / { t/2 } ]^2
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ = \ 1/4 lim_{ t rarr 0 } [ sin( t/2 ) / { t/2 } ]^2
color{blue}{ "now use the Fundamental Trig Limit:" \quad lim_{ x rarr 0 } sin(x)/x \ = \ 1; }
color{blue}{ \qquad "continuing:" }
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ = \ 1/4 cdot [ 1 ]^2
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ = \ 1/4 cdot 1
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ = \ 1/4.
"This is our answer:"
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad lim_{ t rarr 0 } ( 1/t^2 ) sin^2( t/2 ) \ = \ 1/4.