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

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Answer 1 · 2017-08-15T18:23:58

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

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 · 2017-08-15T18:53:39

# f(1)=2/pi.#

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!