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.

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