Homework Help

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

user profile pic

user8235304 | Student, Grade 11 | Valedictorian

Posted July 2, 2013 at 2:46 PM via web

dislike 2 like

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

2 Answers | Add Yours

user profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted July 2, 2013 at 2:59 PM (Answer #1)

dislike 1 like

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`

Top Answer

user profile pic

aruv | High School Teacher | Valedictorian

Posted July 3, 2013 at 7:18 AM (Answer #2)

dislike 1 like

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`

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes