How do you find the limit of #\lim _ { x \rightarrow \pi / 2} \frac { \sin 2x } { \cos x }#?

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Answer 1 · 2017-09-21T12:04:13

The limit is #=2#

The limit is

#lim_(x->pi/2)sin(2x)/cosx=sin(2*pi/2)/cos(pi/2)=sinpi/cos(pi/2)=0/0#

#(1)# This is undefined, we have #2# options, firstly, so we apply L'Hôpital's Rule

#lim_(x->pi/2)sin(2x)/cosx=lim_(x->>pi/2)((sin(2x))')/((cos(x))')#

#=lim_(x->>pi/2)(2cos2x)/(-sinx)=2*cos(pi)/(-sin(pi/2))=-2/-1=2#

and #(2)#, perform some simplifications

#sin2x=2sinxcosx#

#sin2x/cosx=(2sinxcosx)/cosx=2sinx#

Therefore,

#lim_(x->pi/2)sin(2x)/cosx=lim_(x->pi/2)2sinx=2sin(pi/2)=2#