Question

If the function `f(x) = (1 - x)tan(pix/2)` is continuous at `x = 1`, then `f(1)` is ??

Answer 1

The function `f` is not defined at `1` so it is not continuous at `x=1`, but see below.

Explanation:

If we want the function

`f(x) = {((1-x)tan((pix)/2)," ",x != 1),(k," ",x=1):}`

to be continuous, we can find `f(1) = k` to make `f` continuous at `x=1`.

The requirement for continuous at `x=1` is

`lim_(xrarr1)f(x) = f(1)`.

`lim_(xrarr1)f(x) = lim_(xrarr1)(1-x)tan((pix)/2)`

`= lim_(xrarr1)sin((pix)/2)(1-x)/cos((pix)/2)`

`= lim_(xrarr1)sin((pix)/2) lim_(xrarr1)(1-x)/cos((pix)/2)`

The first limit is `1` and the second has form `0/0` so we'll use l'Hospital's Rule

`= (1) lim_(xrarr1)(-1)/(-pi/2 sin((pix)/2))`

`= 2/pi`

So, we need `f(1) = 2/pi`

Answer 2

`f(1)=2/pi.`

Explanation:

For the given fun. `f` to be cont. at `x=1,` we must have,

`lim_(x to 1)f(x)=f(1).`

`:. f(1)=lim_(x to 1) (1-x)tan (pix/2).........(ast).`

Let, `x=1+h, i.e., (x-1)=h," so that, as "x to 1, h to 0.`

`:., by (ast), f(1)=lim_(h to 0) {-h*tan(pi/2(1+h))},`

`=lim_(h to 0){-h*tan(pi/2+pi/2h)},`

`=lim_(h to 0){(-h)(-cot(pi/2h))},`

`=lim_(h to 0) hcot(pi/2h),`

`=lim_(h to 0) (h/sin(pi/2h))*cos (pi/2h),`

`=lim_(h to 0){(pi/2h)/sin(pi/2h)*(2/pi)}cos(pi/2h),`

`={1*(2/pi)}cos0....[because, lim_(y to o) y/siny=1],`

`rArr f(1)=2/pi,` as Respected Jim H. Sir has shown!