How to evaluate the limit of lim(1/t^2)sin^2(t/2) when t approaches 0?

lim(1/t^2)sin^2(t/2)

1 Answer
Mar 6, 2018

"The answer is:" \qquad \quad \quad \ lim_{ t rarr 0 } ( 1/t^2 ) sin^2( t/2 ) \ = \ 1/4.

Explanation:

"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.