How do you evaluate the limit #(sinxcosx)/x# as x approaches #0#?

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Answer 1 · 2018-05-22T21:30:18

#color(blue)(1)#

#(sinxcosx)/x#

Divide numerator and denominator by #x#:

#(sinx/xcosx)/(x/x)#

Cancelling:

#(sinx/xcosx)/(1)#

#lim_(x->0)((sinx)/xcosx)=lim_(x->0)((sinx)/x)*lim_(x->0)(cosx)#

#lim_(x->0)((sinx)/x)=1#

#lim_(x->0)(cosx)=1#

#1xx1=1#

#:.#

#lim_(x->0)((sinxcosx)/x)=1#