How do you find the limit of #sin^2(3t)/t^2# as #t->0#?

1 Answer
Steve M
Nov 9, 2016

# lim_(t rarr 0) sin^2(3t)/t^2 = 9 #

Explanation:

graph{((sin(3x))/x) * ((sin(3x))/x) [-9 9, -0.5, 10.5]}

# lim_(t rarr 0) sin^2(3t)/t^2 = lim_(t rarr 0) (sin(3t)/t)^2 #
# :. lim_(t rarr 0) sin^2(3t)/t^2 = (lim_(t rarr 0) sin(3t)/t)^2 #
# :. lim_(t rarr 0) sin^2(3t)/t^2 = (lim_(t rarr 0) 3/3sin(3t)/t)^2 #
# :. lim_(t rarr 0) sin^2(3t)/t^2 = (3lim_(t rarr 0) 1/3sin(3t)/t)^2 #
# :. lim_(t rarr 0) sin^2(3t)/t^2 = (3)^2(lim_(t rarr 0) sin(3t)/(3t))^2 #
# :. lim_(t rarr 0) sin^2(3t)/t^2 = 9 (lim_(t rarr 0) sin(3t)/(3t))^2 #

Let # theta=3t#, Then As #t rarr 0 => theta -> 0#, So:

# :. lim_(t rarr 0) sin^2(3t)/t^2 = 9 (lim_(theta rarr 0) sin(theta)/(theta))^2 #

And, #lim_(theta rarr 0) sin(theta)/(theta) =1 #is a standard calculus limit.

Hence,

# :. lim_(t rarr 0) sin^2(3t)/t^2 = 9 (1)^2 = 9 #

Which is consistent with the above graph.

Related questions
See all questions in Determining Limits Algebraically
Impact of this question
6243 views around the world
You can reuse this answer
Creative Commons License