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.

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#
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Answer 2

# 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!
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Answer 3

If the function ( f(x) = (1 - x)\tan\left(\frac{\pi x}{2}\right) ) is continuous at ( x = 1 ), then ( f(1) ) exists and is equal to the limit of the function as ( x ) approaches 1.

To find ( f(1) ), we substitute ( x = 1 ) into the function:

[ f(1) = (1 - 1)\tan\left(\frac{\pi}{2}\right) ]

Since ( \tan\left(\frac{\pi}{2}\right) ) is undefined, we need to evaluate the limit of the function as ( x ) approaches 1 to determine if it is continuous at ( x = 1 ).

[ \lim_{x \to 1} (1 - x)\tan\left(\frac{\pi x}{2}\right) ]

To evaluate this limit, we can use the properties of limits. Since ( \tan\left(\frac{\pi}{2}\right) ) is undefined, we need to rewrite the expression.

We know that as ( x ) approaches 1, ( \frac{\pi x}{2} ) approaches ( \frac{\pi}{2} ).

Therefore, we rewrite the expression as:

[ \lim_{x \to 1} (1 - x)\tan\left(\frac{\pi}{2}\right) ]

Now, we can directly substitute ( x = 1 ) into the expression:

[ (1 - 1)\tan\left(\frac{\pi}{2}\right) = 0 ]

So, if the function ( f(x) = (1 - x)\tan\left(\frac{\pi x}{2}\right) ) is continuous at ( x = 1 ), then ( f(1) = 0 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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