How do you find the limit of #(sin²x)/(x)# as x approaches 0?

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Answer 1 · 2016-08-02T00:31:42

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#lim_(x to 0) (sin²x)/(x)#

#lim_(x to 0) (sinx)/(x) * sin x#

#lim_(x to 0) (sinx)/(x) * lim_(x to 0) sin x#, we can seperate because sin x is continuous through the limit

# = 1 * 0 = 0 #

because #lim_(x to 0) (sinx)/(x) = 1# is a very well known limit

OR

#lim_(x to 0) (sin²x)/(x)#

is #0/0# indeterminate, so we can use L'Hopital

#= lim_(x to 0) (2 sinx cos x)/(1)#

#= lim_(x to 0) sin 2x = 0#