How do you make a function that has an asymptotic/infinite discontinuity continuous?

Question

Find the value of k that would make the function continuous in each case:

h(x) = sin[(5*pi*x) -1]/(2x-1), x is not equal to 1/2
k, x =1/2

that was a piecewise function, by the way :p
thank you

ted s · Accepted Answer

k = 0 and h(x) = [ sin(5πx ) - 1 ] / [ 2x - 1 ] , x╪ 1/2

if you did type correctly then k is not finite

ortiz · Answer

For x greater desirable than 3, the function does not have a real value. On one area of three (such as 3), the function has actual values. on the different area, it has complicated values. yet ... the value on the two sides of three is an identical. it is, f(3-eps) = f(3 + eps), for infinitely small value of eps. And in case you graph the fee of f(x), you will see that it is non-stop. So i might say that f(x) is discontinuous on the actual numbers, yet non-stop on the complicated numbers.

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How do you make a function that has an asymptotic/infinite discontinuity continuous?

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