10 points to the first person to help me with continuities!?

Question

Show that f(x) = {(x^2 - x - 2)/(x + 1) , x < -1

{ 2x +2 , x>_ -1

is not continuous at x = -1 but is continuous from the right of x = -1

Show that f(x) = (x-1)/(x(x+1)) is not continuous at x = 0 or x = -1 and show also that these discontinuities are nonremovable.

Answer 1

basically at x=-1 the value of f(x) is 2(-1)+2 which is 0 but left of negative one you have f(x) with a denominator of (x+1) and remember when the denominator is 0 the function is undefined. so as x is getting arbitrarily close to -1 the value is undefined. try graphing it out and you will see.

same thing for this one when the denominator is 0 the function is undefined and thus not continuous. the denominator is  (x(x+1) so if x =0, (0(0+1)=0

and if its equals -1

(-1(-1+1))=0 and it becomes undefined

by non-removable it means the denominator can be factored out.

Lets say you have (x+1)(x+3)/(x+1) the denominator is  0 at x=-1  but the numerator also has (x+1) so it can be factored out.

(x-1)/(x(x+1)) cannot be factored  

Answer 2

OK,

1.) if we put -1 in f(x) then f(x) becomes 0 / 0 which is undefined and cannot be plotted on the graph, therefore there is a break in the graph and hence -1 is the point of discontinuity but at the right of -1, f(x) is just a straight line [factorize x^2 - x - 2, it will become (x - 2) ( x + 1)], and therefore continuous.

Note that you cannot cancel the factors (x + 1) when discussing the continuity as you have to take the domain of the function correctly.

2.) Same reason, if you try to put 0 or -1 then the denominator will become 0 and division by 0 is not possible. Therefore the value of the function on both the points 0, -1 cannot be plotted on the graph and hence discontinous function at 0 and -1.

Also note that if factors are cancelling then discontinuity is removable as in 1.) but if they are not like in 2.) then discontinuity is non-removable.

Answer 3

For x&lt;-1, f(x)=(x^2 - x - 2)/(x + 1)=(x-2)

lim(x-&gt;-1) (x-2)=-1-2=-3

But f(-1)=-2+2=0, hence discontinuous at x=-1.

lim(x-&gt;-1) (2x+2) = -2+2=0, which is equal to f(-1), hence continuous from the right of x=-1.

For the next problem. f(0) and f(-1) are undefined, hence f(x) is discontinuous at these 2 points.

f(0-)-&gt;+inf but f(0+)-&gt;-inf, hence it is unremovable at x=0

Similarly, f(-1-)-&gt;-inf but f(-1+)-&gt;+inf, hence it is also unremovable at x=-1.