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
martin23239
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. #

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