#lim_(xrarrpi/2) 1-sinx/(x-π/2)#?

Question

A. -2
B. -1
C. 1
D. 0
E. 2

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Answer 1 · 2018-03-23T23:11:15

If we set #u=x-pi/2# hence #x=u+pi/2# we have that

#lim (1-(sin(u+pi/2)/u))=lim(1-cosu/u)=lim((u-cosu)/u)#

Because #u->0# and #u-cosu->-1# we have that the limit is

#lim_(u->0) ((u-cosu)/u)->oo#

Hence none of the given answers is correct.

If you mean the limit #lim (1-sinx)/(x-pi/2)# then apply D'Hopital law to get

#lim_(x->pi/2) ((1-sinx)')/((x-pi/2)')=lim_(x->pi/2)(-cosx)=-cos(pi/2)=0#

Hence the correct answer is D. #0#