To 6x squared plus 18x plus 12 over x squared minus As a corollary, if every point of U is an isolated point of U, then ƒ is continuous on U.Īs a result, we arrive at the following (counterintuitive?) result: since any sequence of real numbers is a real-valued function defined on the set of natural numbers, and since every natural number is an isolated point of the set of natural numbers (can you prove this? Hint: take r = 1/2 in the definition above), every sequence of numbers is a continuous function! (This fact is mainly a result of extreme cases of the definition of continuity.) If u is an isolated point of U, then ƒ is continuous at u (this may be proved). Suppose now that ƒ: U → R is a real-valued function defined on U. We say that a point u in U is an isolated point of U if and only if there exists a real number r > 0 such that the open interval (u - r, u + r) contains no point of U other than u in other words, u is an isolated point of U if there exists some open interval about u which contains no other points of U. Suppose U is any non-empty subset of the real numbers, R. This is perhaps a bit more abstract than what you have in mind, but the property of being continuous at the only point of definition may be somewhat extended as follows: Yes, a function defined at only one point, is continuous at that point. Some are talking about some sort of hospital □ rule, I haven't learnt that yet so sorry if I'm wrong So it's not possible to make the function continuous here. However, when you input x=+2 to the original equation, you get 72/0 which shows that the graph is curving up towards infinity here at x=+2. We say the limit as x approaches -2 of f(x) = 1.5 It would seem to "make sense" to assign a number to that question mark in between these,īy making x really close to -2 you can get the corresponding y value to be as close as 1.5 as your efforts in making x close to -2. I don't know how to show graphs here, but here are a few values: This is meaningless, and if you plot the equation on a graphing software you'll see the graph on either side of this point is quite smooth, apart from just one hole here at x=-2. Notice that, in the original equation, inputting x = -2 will evaluate to 0/0. I haven't learnt much, but this is what I understand:
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