# what type of discontinuity?

• Jan 27th 2010, 12:47 PM
haebinpark
what type of discontinuity?
f(x)= (e^x) / (4 - x^2) x <,= 0
sin(pi/2 - x) / (x+2)^2 x>0

i found that f(0)=LHL=RHL=1/4
thus, f(x) is continuous at x=0

however, there's another question that ask "locate, with justification, any points of discontinuity"

i figured that when x= -2, it became discontinuity,
however, what kind of discontinuity is it?

f(-2) = LHL = e^-2 / 0
RHL = 3/0

is this removable discontinuity?
• Jan 27th 2010, 01:24 PM
drumist
If the limit from either side goes to positive or negative infinity, then it is called an infinite discontinuity.

Basically, three types:

If either side goes to positive or negative infinity, it is an infinite discontinuity.

If both sides are finite and equal, it is a removable discontinuity.

If both sides are finite but not equal, it is a jump discontinuity.
• Jan 27th 2010, 01:56 PM
haebinpark
but this doesn't meet any of those statement...?
• Jan 27th 2010, 02:45 PM
skeeter
Quote:

Originally Posted by haebinpark
but this doesn't meet any of those statement...?

yes it does ... as $\displaystyle x \to -2^-$ , $\displaystyle f(x) \to -\infty$

as $\displaystyle x \to -2^+$ , $\displaystyle f(x) \to \infty$
• Jan 27th 2010, 02:50 PM
haebinpark
how's that work?

when x is approaches from the left

lim x->-2- e^x/(4-x^2)
which is +/0 and is +inf

when x is approaches from the right

lim x->-2+ [sin (pi/2)-x]/(x+2)^2 = 3/0
which also is +/0 +inf ?

am i wrong?
• Jan 27th 2010, 03:03 PM
Plato
But against forum rules you deleted the original post.
That may earn you an infraction.

So here it is again. That function is not even defined for $\displaystyle x=-2$ so there is no way for it to be continuous there.
• Jan 27th 2010, 04:38 PM
haebinpark
okay... this is really confusing... :s

so, "skeeter" is saying that it is jump discontinuity

and "plato" is saying the function is not even defined...

umm..

but i still don't see how Skeeter came up with that answer

i still get

when x is approaches from the left

lim x->-2- e^x/(4-x^2)
which is +/0 and is +inf

when x is approaches from the right

lim x->-2+ [sin (pi/2)-x]/(x+2)^2 = 3/0
which also is +/0 +inf ?

removable...? jump...?

i graphed on calculator and it seems like... removable.... but i am not sure...
• Jan 28th 2010, 03:50 AM
haebinpark
haaaaaaaaa
can someone please tell me where i went wrong?
• Jan 28th 2010, 04:04 AM
skeeter
Quote:

Originally Posted by haebinpark
haaaaaaaaa
can someone please tell me where i went wrong?

you need to take a look at how f(x) is defined at for values of x < 0

you're using the wrong part of the function.
• Jan 28th 2010, 04:21 AM
HallsofIvy
Quote:

Originally Posted by haebinpark
okay... this is really confusing... :s

so, "skeeter" is saying that it is jump discontinuity

and "plato" is saying the function is not even defined...

umm..

Skeeter did NOT say it was a jump discontinuity. drumist told you "If both sides are finite but not equal, it is a jump discontinuity. "

The limits from the two sides are not finite.

However, plato saying "the function is not even defined" did not itself mean it was not a jump discontinuity- there are different ways of being "not defined". For example, the function, f(x)= -1 if x< 0 and f(x)= 1 if x> 0 is not defined at x= 0. The limits from left and right are -1 and 1 so that is a jump discontinuity.

Quote:

but i still don't see how Skeeter came up with that answer

i still get

when x is approaches from the left

lim x->-2- e^x/(4-x^2)
which is +/0 and is +inf
No, that's wrong. If, for example, x= -2.01, $\displaystyle \frac{e^{-2.01}}{4- (-2.01)^2}= \frac{0.134}{4- 4.0804}= \frac{0.134}{-.0804}$ which is clearly negative.

Quote:

when x is approaches from the right

lim x->-2+ [sin (pi/2)-x]/(x+2)^2 = 3/0
which also is +/0 +inf ?

removable...? jump...?

i graphed on calculator and it seems like... removable.... but i am not sure...
Your error is thinking that "+/0" is positive infinity. It is not, of course, any number and whether the limit is positive or negative infinity depends upon what the denominator is like close to the limit point.

I graphed it on a calculator myself. I don't see how you could say it looked like a removable discontinuity! Approaching -2 from below the graph goes off toward -infinity, approaching form above, toward infinity. Even if both sides went to infinity, that would not give a "removable discontinuity". To have a removable discontinuity, the two sides must approach the same finite value, not the value of the function, so that you could "remove" the discontinuity by redefining the function at that point.