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Precalculus

Course: Precalculus>Unit 10

Lesson 9: Exploring types of discontinuities

Types of discontinuities

A function being continuous at a point means that the two-sided limit at that point exists and is equal to the function's value. Point/removable discontinuity is when the two-sided limit exists, but isn't equal to the function's value. Jump discontinuity is when the two-sided limit doesn't exist because the one-sided limits aren't equal. Asymptotic/infinite discontinuity is when the two-sided limit doesn't exist because it's unbounded.

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• Is an asymptotic discontinuity any different than an infinite discontinuity?
• They are the same thing – if you look in the `About` text it actually says "Asymptotic/infinite discontinuity".
• I understand that classification of discontinuities is 3 types
i. point removable
ii. jump discontinuities
iii.asympotic discontinuities

anything also
• There are also oscillating discontinuities. Look at the graph of f(x)=sin(1/x). It has no value or limit at x=0.
• I can't understand why the value of the y=x^2 graph at x=3 is 4, and not 9. Probably an obvious answer, but it's eluding me!
• So can you see the dot that is separated from the curve?
• Hi, I am learning how to evaluate functions by direct substitution right now. I was wondering why simply substituting or re-arranging a function would automatically give us the limit at that point. For example, in the two graphs on the left in this video, the y-value is defined at the x-value but the limit either doesn't equal that same y-value or doesn't exist. I want to see the actual functions that could result in these two graphs to better understand why we can directly substitute without fear of scenarios like these two. Is it only possible for piece-wise functions to create these types of scenarios? I feel like I am overlooking something and would really appreciate the help. Thanks in advance!
• A function can be determined by direct substitution if and only if lim_(x->c)_ f(x) = f(c). In other words, as long as the function is not discontinuous, you can find the limit by direct substitution.

There is also another way to find the limit at another point, and that is by looking for a determinant for the indeterminate form by using other methods and defining it by using another function. For example, lim_(x->2) (x^2 + 4 x - 12)/(x - 2), determined directly, equals (0/0), indeterminant form. However, there are many ways to determine a function by simply simplifying the function when direct substitution yields the indeterminant form. For this example, you could simply factor the limit to get lim_(x->2)_ (x+6), x ≠ 2. The constraint is added to be mathematically correct when it comes to being equivalent to the limit beforehand. However, say you found a function that is similar to the simplified function, only without the constraint, called g(x) = (x+6). You can define that as your new limit: lim_(x->2) g(x) = 8, thus lim_(x->2) (x^2 + 4 x - 12)/(x - 2) = 8.

• Is a quadractic formula discontinous
• Well, the quadratic formula is a formula, so it can't be graphed.

However, a function related to the quadratic formula, a quadratic polynomial, is continuous over its entire interval.
• What about the function which has one sided limit? For example if lim_(x->0) when approaching from the right exists, but lim_(x->0) when approaching from the left is asymptote ?
Does this qualify as asymptotic discontinuity or some kind of mix of jump/asymptotic discontinuity?
• That would be an asymptotic discontinuity. Asymptotic discontinuities are defined as occurring when at least one of the one-sided limits are undefined.
• Do sharp peaks or turns in the line, such as the ones in some parametric equations, count as a discontinuity?
• The derivative of said functions would be discontinuous, but as long as the line never breaks it is continuous. Still, sharp turns or other sudden changes in slope will make the function non differentiable. So still something you have to keep an eye out for.