- #1
Rodrae
- 13
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/x/ = y is non removable discontinuity
x
but is
/x/^2 = y is a removable discontinuity or not? and why?
x^2
x
but is
/x/^2 = y is a removable discontinuity or not? and why?
x^2
Rodrae said:/x/ = y is non removable discontinuity
xbut is
/x/^2 = y is a removable discontinuity or not? and why?
x^2
1/|x| is discontinuity at zero and non removable... because we couldn't simplify the equation to remove x w/c is 0agentredlum said:You can tell from the graphs of these functions whether the discontinuity is removable or not.
|x|/x is discontinuous for sure at x = 0
For positive x the graph is the horizontal line y = 1
For negative x the graph is the horizontal line y = -1
This causes a 'jump' discontinuity at x = 0 the y value jumps from -1 to 1 at x = 0
'Jump' discontinuities are not removable because you can not make the graph continuous by filling in a SINGLE point where the problem occurs. In this case at x = 0
Infinite discontinuities are also non-removable, because you can not make the graph continuous by filling in a SINGLE point.
|x^2|/x^2 is discontinuous for sure at x = 0
however graph of |x^2| is exactly the same as graph of x^2 so absolute value symbol is un-necessary since extra branch is not introduced.
So considering values very close to 0, on the left of 0 |x^2|/x^2 gives 1, on the right of zero |x^2|/x^2 gives 1. This is not a 'jump' discontinuity (or an infinite discontinuity) and the function can be made continuous by filling in a SINGLE point.
So |x^2|/x^2 has a removable discontinuity at x= 0
The key is a discontinuity is removable if you can 'fix' discontinuity using a single point.
Heres a bonus question...1/|x| ... is the discontinuity at x= 0 removable or non removable and why?
Rodrae said:1/|x| is discontinuity at zero and non removable... because we couldn't simplify the equation to remove x w/c is 0
The graph will move closer but never reach x=0
alexfloo said:That all sounds about correct. A rational function before and after you cancel out those factors are not technically the same function. This is because they have a different domain. (The first one wasn't defined where the denominator had a zero.) However, everywhere that those two, different, functions are defined, they're equal to each other. When you cancel, you find a different function that is *almost* exactly the same as the old one, except that it doesn't have the point of discontinuity. So, you've removed it. Therefore, it must've been a removable discontinuity.
agentredlum said:Yes, very good. Notice that this discontinuity is different than the discontinuities of |x|/x and |x^2|/x^2
For 1/|x|, very close to zero, from the right graph goes to positive infinity as x goes to zero, from the left graph goes to positive infinity as x goes to zero. Infinity is not a number, therefore we cannot 'fill it in' using a single point. Even though the function appears to approach the same place from both sides of zero, the place must be a well defined number otherwise it has no limit, as we say. Limits are very important in continuity.
Bonus question to the bonus question...
y = 1/x has a non-removable discontinuity at x = 0, how is this non-removable discontinuity different than the non-removable discontinuity of y = 1/|x|
Hint* Look at both graphs and ask yourself what is happenning to the y values as x approaches 0 from both sides.
http://www.wolframalpha.com/input/?i=plot+1/|x|&asynchronous=false&equal=Submit
http://www.wolframalpha.com/input/?i=plot+1/x&asynchronous=false&equal=Submit
Rodrae said:In 1/x
y can be negative if x is negative and goes closer and closer but will never reach the graph y=0 and x=0
but in 1/|x|
y will never be a negative no because |x| is always positive the graph goes closer and closer but will never reach to x=0 and y=0
agentredlum said:Your statements are correct. You are explaining the behavior of the entire graphs. However, when looking at a discontinuity of a single point, it is not a good idea to look at the entire graph. Your interest should be only for what happens at x = 0 because that is where the problem occurs and more subtle analysis is needed for understanding the behavior at x = 0
The answer i was looking for is 1/|x| goes to positive infinity from both sides of x = 0
1/x goes to negative infinity on the left of x = 0, positive infinity on the right of x = 0
The example was meant to show 2 different kind of infinite non-removable discontinuities.
A removable discontinuity of /x/ is a type of discontinuity that occurs when a function has a hole or gap in its graph at a specific value of x. This means that the function is undefined at that point, but the limit as x approaches that value exists. It can also be referred to as a point discontinuity.
If a function has a removable discontinuity of /x/, it will have a hole or gap in its graph at a specific value of x. This can be observed by graphing the function or by looking at the function's equation and identifying any values of x that cause a division by zero error.
A removable discontinuity of /x/ is caused by a factor in the function's equation that can be canceled out, resulting in a hole or gap in the graph. This commonly occurs when the denominator of a rational function is equal to zero, causing a division by zero error.
Yes, a removable discontinuity of /x/ can be removed by filling in the gap in the graph with the limit of the function as x approaches the value of the discontinuity. This will result in a continuous function with no holes or gaps.
The main difference between a removable discontinuity of /x/ and a non-removable discontinuity is that the limit of the function exists at a removable discontinuity, but does not exist at a non-removable discontinuity. This means that a removable discontinuity can be removed, while a non-removable discontinuity cannot be removed and will always result in a discontinuous function.