calculus please help!
posted by Terry on .
Which of the following functions f has a removable discontinuity at a? If the discontinuity is removable, find a function g that agrees with f for x a and is continuous at a. (If the discontinuity is not removable, enter NONE.)
1. f(x)= (x^41)/(x1), a=1
?The discontinuity is removable.
?The discontinuity is not removable.
g(x) = ?
2. f(x)= (x^3x^212x)/((x4), a=4
?The discontinuity is removable.
?The discontinuity is not removable.
g(x) = ?
3) f(x) = [[ sin(x) ]], a = ð
(Recall that [[h(x)]] means the largest integer that is less than or equal to h(x).)
?The discontinuity is removable.
?The discontinuity is not removable.
g(x) =?

I don't understand what they are asking! If anyone could explain that, that would be a great help!

A removable discontinuity is a point where the function does not exist, but it's limit exists on both sides for an interior point, and on the interior side if it is an end point.
The discontinuity can be removed by defining the function at the point of discontinuity as the limit.
Take for an example the first question:
f(x)=(x^41)/(x1)
Since
x^41 = (x²+1)(x+1)(x1)
it is evident that the limit at x=1 exists on both sides, but f(x) is undefined at x=1.
By redefining f(x) as
g(x)=(x^41)/(x1) for x≠1, and
g(x)=4 for x=1
g(x) is now continuous on ℝ, and the discontinuity has been removed in the new function.
You can work on the other problems on this basis.