How to evaluate this limit?

Question

#lim x>0 (sinx-sin ^2x)/x#

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Answer 1 · 2018-05-08T19:15:38

# 1#.

We will use : #lim_(theta to 0)sintheta/theta=1#

#lim_(x to 0)(sinx-sin^2x)/x#,

#=lim_(x to 0){sinx/x-sin^2x/x}#,

#=lim_(x to 0){sinx/x-(sinx/x)sinx}#,

#={1-(1)sin0}#,

#=1-0#,

#1#.