How do you find the limit of #(sin^2(4t)) / t^2 # as t approaches 0?

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Answer 1 · 2016-05-16T16:15:10

Use algebra, properties of limits and the important limit #lim_(xrarr0)sinx/x=1#

#sin^2(4t)/t^2 = (sin(4t)/t)^2# (Algebra)

So we need only find #lim_(trarr0)sin(4t)/t# and square that result. (Property of limits)

There are 3 big tricks in mathematics:

In this case we'll use item 2. to accomplish 3.

#sin(4t)/t = 4/4 sin(4t)/t = 4(sin(4t)/((4t)))#

Now, as #trarr0#, we also see that #4trarr0#, so thinking of #4t# as #x# in #lim_(xrarr0)sinx/x=1#,

we arrive at #lim_(trarr0)(sin(4t)/((4t)))=1#

So #lim_(trarr0)4(sin(4t)/((4t))) = 4(1) = 4#

And finally

#(lim_(trarr0)sin(4t)/t)^2 = (lim_(trarr0)4(sin(4t)/4t))^2 #

# = (4)^2 = 16#