Evaluate lim(x tends to 1) cosπ/2x/1-x?

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Answer 1 · 2018-03-10T08:55:23

#lim_(x->1) cos(pi/2x)/(1-x)=pi/2#

As in #lim_(x->1) cos(pi/2x)/(1-x)#,

as #x->1#, both numerator and denominator tend to #0#.

Hence, we should use L'Hospital rule, according to which

#lim_(x->1)(f(x))/(g(x))=lim_(x->1)((df)/(dx))/((dg)/(dx))#

Hence #lim_(x->1)(-sin(pi/2x)*pi/2)/(-1)#

= #pi/2lim_(x->1)sin(pi/2x)#

= #pi/2*sin(pi/2)#

= #pi/2#