Discontinuity

[dear2009]

Dear users of mathhelp

Can you please help with these two problems, I just cant seem to fully grasp the whole problem

in this problem y = (x-2)/(x-3)

i wanted to know is this removable or non-removable

i believe it is removable discontinuity but I am not fully sure why

Another problem I have is this
g(x) = 2/3x, [x, x + h]


I got (2) / [3 (x + h) x]


Thanks in advance.

 

[Deleted member 4993]

Dear users of mathhelp

Can you please help with these two problems, I just cant seem to fully grasp the whole problem

in this problem y = (x-2)/(x-3) <<< Non -removable

Removable discontinuity means that we can "simplify" the function in such way that the discontinuity will be removed

suppose

f(x) = (x[sup:15zzvh5n]2[/sup:15zzvh5n] - 4)/(x-2)

f(x) has a removable discontinuity at x = 2 - because:

f(x) = (x[sup:15zzvh5n]2[/sup:15zzvh5n] - 4)/(x-2) = [(x-2)(x+2)]/(x-2) = x+2 <<<< Discontinuity is removed.
i wanted to know is this removable or non-removable

i believe it is removable discontinuity but I am not fully sure why

Another problem I have is this
g(x) = 2/3x, [x, x + h]


I got (2) / [3 (x + h) x]


Thanks in advance.

 

[BigGlenntheHeavy]

No, no, Subhotosh Khan.


$\displaystyle f(x) \ = \ \frac{x^{2}-4}{x-2} \ = \ \frac{(x+2)(x-2)}{x-2} \ = \ x+2 \ with \ restriction \ x \ \ne2, \ a \ discontinuity.$

$\displaystyle However, \ to \ remove \ discontinuity, \ f(x) \ = \ \frac{x^{2}-4}{x-2} \ = \ x+2, \ with \ restriction \ f(2) = 4$

$\displaystyle f(x) \ = \ \frac{x^{2}-4}{x-2} \ \ne \ f(x) \ = \ x+2.$

 

[Deleted member 4993]

Yes ... Yes .... Yes.. Glen

The definition of removable discontinuity is :

The first way that a function can fail to be continuous at a point a is that


lim f(x) = L exists (and is finite)
x --> a

but f(a) is not defined or f(a) <> L.

Discontinuities for which the limit of f(x) exists and is finite are called removable discontinuities

for f(x) = (x[sup:8yxxhaap]2[/sup:8yxxhaap]-4)/(x-2) the limit exists (= 4) and it is finite - thus it is removable.

For the function is question f(x) = (x-2)/(x-3) the limit is not finite - and hence non-removable.