# What is`lim_(x->1) ( 1 - x)*(tan ((pi*x)/2))` = ?

The limit `lim_(x-> 1)(1 - x)*tan ((x*pi)/2)` has to be determined.

`lim_(x-> 1)(1 - x)*tan ((x*pi)/2)`

= `lim_(x-> 1)((1 - x)*sin((x*pi)/2))/cos((x*pi)/2)`

Substitute x = 1

=> `((1 - 1)*1)/0 = 0/0`

As this is in an indeterminate form `0/0` , use l'Hopital's rule and substitute the numerator and denominator with their derivatives.

=> `lim_(x-> 1) (-1*sin(x*pi/2) + (1 - x)*(pi/2)*cos(x*pi/2))/((-pi/2)*sin (x*pi)/2)`

Substitute x = 1

=> `-sin(pi/2)/(-pi/2*sin(pi/2))`

=> `2/pi `

The limit `lim_(x-> 1)(1 - x)*tan ((x*pi)/2) = 2/pi`

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