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

### 2 Answers | Add Yours

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`

Let x-1=y

as x->1 ,y->0

Thus

`lim_(x->1)(1-x)tan((pix)/2)=lim_(y->0)(-y)tan((pi/2)(y+1))`

`=lim_(y->0)(-y)(-cot((piy)/2))`

`=lim_(y->0)((piy)/2)/(pi/2){cos((piy)/2)/sin((piy)/2)}`

`=(2/pi)lim_(y->0){((piy)/2)/sin((piy)/2)}lim_(y->0)cos((piy)/2)`

`=(2/pi)`

`` because of

`lim_(x->0)sin(x)/x=1`

`lim_(x->0)cos(x)=1`