# Find the x-values at which f is not continuous. Are these discontinuities removable?Find the x-values at which f is not continuous. Are these discontinuities removable.(Use k as an arbitrary...

Find the x-values at which f is not continuous. Are these discontinuities removable?

Find the x-values at which f is not continuous. Are these discontinuities removable.(Use k as an arbitrary integer if needed).

f(x) = tan* (πx)/(2)

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should check if the tangent function is continuous, hence, you need to check if `lim_(x->1, x<1) tan(pi*x/2) = lim_(x->1, x>1) tan(pi*x/2)`  such that:

`lim_(x->1, x<1) tan(pi*x/2) = +oo`

`lim_(x->1, x<1) tan(pi*x/2) = -oo`

Notice that the side limits have no equal values and the function has a vertical asymptote at `x=1` , hence, the function `f(x) = tan(pi*x/2)` is discontinuous at `x = 1` .

quantatanu | Student, Undergraduate | (Level 1) Valedictorian

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f(x) is discontinuous not only at x = 1 but at

x = 2*k + 1

for k = 0, 1, 2, 3,...

It was indicated as a hint in the question also

"(Use k as an arbitrary integer if needed)."

Moreover you have not discussed about whether the discontinuities are of "removable" nature or not !

Check the answer that I have provided at the top, it discusses every part of the problem and also gives the complete answer to the problem.

quantatanu | Student, Undergraduate | (Level 1) Valedictorian

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Given function is:

f(x) = tan[Pi x/2] = sin[Pi x/2] / cos[Pi x/2]

So this function is discontinuous at the points where it blows up means reaches +/- infinity. Its easy to see that f(x) reaches infinity

when

cos[Pi x/2] = 0

=> Pi x/2 = (2*k + 1) Pi/2

=> x = 2*k + 1    [where k is an arbritrary integer, +ve or -ve]

so f(x) is discontinuous at  x = 2*k + 1,

These discontinuities are removable because these points are wellseparated, the separation between two adjacent discontinuity is:

separation = (2 *(k+1) + 1) - (2*k + 1)

=  2

and well separated discontinuity which occur at only distinct points are removable. To understand this concept I should show you a discontinuity which is not removable:

Consider the function, g(x) = exponential[1/x], very near to x = 0 every point makes g(x) infinite and you cannot separate one point from other. These kind of discontinuity are non-removable.