Find the limit when x is approaching cfrom the left or the right. How do you solve a one-sided limit and continuity on a closed interval?
When it gives me the graph of these I can do them because i can see the point that x is approaching. When I have to find the limit when it just gives me the function I don't even know how to begin to solve them. It makes no sense. None of this makes any sense. Even when i know how to solve the limits, I don't really understand what I am doing. What even is the limit of a function? What is continuity? What is a closed interval? On most things I can run through the motions and steps of how to solve a problem but I don't actually know what I am doing. On some of the examples I see there are numbers and I dont even understand where they are coming from. I don't know how to solve a problem when there is no set way to do it or just one way that will work. I understand basic limits because you can just plug in the number x is approaching and find the limit. But now that I am getting into more advanced and I can't do it like that. Now i dont know what I am doing.
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(1) A function is continuous on an interval if for every point c in the interval the limit of the function at c exists and is equal to f(c).
(2) A closed interval includes the endpoints. This is usually denoted [a,b] so that for any point c in the interval `a<=c<=b`
(3) In order to determine continuity on a closed interval the function must be continuous at every point c on the open interval (a,b) (that is a<c<b) and in addition there is the requirement that the limit as x approaches a from the right be f(a) and the limit as x approaches b from the left is f(b)
(4) So you need to be able to evaluate one sided limits. The nice thing is that for "nice" functions you can indeed "plug in" or substitute c into the function. For example, `lim_(x->0^+)sqrt(x)=0` . Note that to ask about the limit from the left is meaningless in the real numbers. (Nice functions include polynomials, trigonometric functions, etc... on their domains)
However, you cannot always substitute. For instance, if you are dealing with infinite limits `lim_(x->0^+)1/x=oo,lim_(x->0^-)=-oo` and `1/0` is undefined. Here you take into account the behavior of the function in the neighborhood of 0.
Another case where you cannot substitute is a piece-wise function with a jump discontinuity. For example let `f(x)=x^2` if `x<=1` and `f(x)=2x-3` if x>1. The limit as x approaches 1 does not exist. But `lim_(x->1^+)=2(1)-3=-1` while `lim_(x->1^-)=1^2=1` . The one-sided limits exist. By definition, a function is continuous at a point if and only if the left and right hand limits exist and agree with the value of the function at that point.
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