Question

Which function has a removable discontinuity?

Answer 1

1. \[\frac{ 5x }{ x-x^2 } \]

Answer 2

2. \[\frac{ x+2 }{ x^2-x-2 }\]

Answer 3

1. it has removal discontinuity by defining x not equal to 1

Answer 4

why?

Answer 5

Removable discontinuities are ones where the limit exists, but the function isn't defined at that point.

Answer 6

You might want to look for *points of discontunity* first. Both of your functions have two of them.

Answer 7

firstly if we want to define function.it should be Continuous through out range
and denominator should not equal izero.
hence we can see Q1 . if we put x=1 function is not define,
hence we can define function except x=1
hence it is removal discountinuties.

Answer 8

Tip #1 @kaylalynn
When you have a rational function (functions that look like fractions) your best bet at finding points of discontinuity are the times when the denominator is zero.

Answer 9

i thought there couldn't be a zero at the denominator

Answer 10

it is at x= 1

Answer 11

That's just it :)
The denominator can't be zero ^.^
SO... for whenever the denominator is zero, the function is discontinuous at that point :)
Here's an example
\[\huge \frac{x+1}{x-1}\]
The denominator is zero when x = 1
Hence, x=1 is a point of discontinuity :)

Answer 12

oh okay. so it is the first 1?

Answer 13

yes

Answer 14

How do you know it's the first one? Both of these functions have points of discontinuity.

Answer 15

well do either of them have removal discontinuity?

Answer 16

i have 4 of them

Answer 17

That's what we're supposed to find out :P
But first, look for the points of discontinuity. When is the denominator of equation 1 equal to zero?

Answer 18

when x = 1?

Answer 19

Yes, and...?

Answer 20

0?

Answer 21

Yes.
So the denominator of the first is zero, when x=0 or x=1, correct?

Answer 22

yes

Answer 23

Now
Tip #2
If you're looking for a removable discontinuity in these rational functions, try to see if for any of the x-values where the denominator is zero, the numerator is also zero.

Answer 24

well 5(0) is 0 right?
so it would be 0/0?

Answer 25

Yes, so that's likely a point of discontinuity that's removable :)
Just out of interest, can you find
\[\huge \lim_{x\rightarrow 0}\frac{5x}{x-x^2}\]?

Answer 26

maybe let me try

Answer 27

i got \[\lim_{x \rightarrow 0} x + x^2 = 0 ?????\]

Answer 28

Here: Study this well, this trick comes in handy many a time :)
\[\huge \lim_{x\rightarrow 0}\frac{5x}{x(1-x)}\]
\[\huge \lim_{x\rightarrow 0}\frac{5\cancel{x}}{\cancel{x}(1-x)}\]
\[\huge \lim_{x\rightarrow 0 }\frac{5}{1-x}=5\]

Answer 29

okay thank you i see what i did wrong

Answer 30

So as you can see, the limit DOES exist when x goes to zero, but the function does not exist when x = 0 (because the denominator becomes zero) Hence the removable discontinuity at x = 0

Answer 31

Now work on the second one... when is the denominator equal to zero in equation (2)?

Answer 32

x = 2?

Answer 33

Yes... and there's another one :)

Answer 34

but its not removable though unless its -2 right?

Answer 35

Yes. :)
You can do it the other way, by the way, in that you can get where the NUMERATOR is zero, and see if the denominator is also zero at that point. It seems easier in the case of equation (2).

Answer 36

So, if there is no value for x which would make BOTH the numerator and denominator zero, then your rational function can't have any removable discontinuity :)

Answer 37

so it is not removable

Answer 38

So there is no removable discontinuity, as the only value that would make the numerator zero is, as you said, x = -2
but this does not make the denominator zero.

Answer 39

okay:)

Answer 40

thanks:)

Answer 41

No problem :)