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- Dictionaryunbounded/ʌnˈbaʊndɪd/
adjective
- 1. having or appearing to have no limits: "the possibilities are unbounded"
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Sep 9, 2015 · Here are four examples... x The simplest example of an unbounded function is f(x) = x, which is unbounded for x in (-oo, oo) 1/x The function f(x) = 1/x is unbounded on any interval that includes x = 0, due to a simple pole at x = 0. tan(x) The function f(x) = tan(x) is unbounded on any interval that includes an x of the form pi/2 + npi, since it has a vertical asymptote at each of these values.
If the function is unbounded, the graph would progress to infinity, in some direction(s). A. S. Adikesavan · 1 · Mar 2 2016
Oct 27, 2014 · A function f is bounded in a subset U of its domain if there exist constants M, m in RR such that m<=f(x)<=M, for all x in U.
Apr 25, 2017 · See explanation. Definitions: A set is bounded above by the number A if the number A is higher than or equal to all elements of the set. A set is bounded below by the number B if the number B is lower than or equal to all elements of the set. Examples: Example 1 A set of natural numbers NN is bounded below by the number 0 or any negative number because for all natural numbers n we have: 0<=n and for every negative number N we have N<=n Example 2 Let A be a set A={1/n: n in NN} This set can ...
Sep 24, 2015 · The feasible region is the set of all points whose coordinates satisfy the constraints of a problem. For example, for constraints: x >= 0, y >= 0, x+y <= 6, y <= x+3 The feasible region is shown below. (Desmos graphing utility.)
Sep 18, 2015 · You tend to see "undefined" when dividing by zero, because how can you separate a group of things into zero partitions? In other words, if you had a cookie, you know how to divide it into two parts---break it in half. You know how to divide it into one part---you do nothing. How would you divide it into no parts? It's undefined. 1/0 = "undefined" You tend to see "does not exist" when you encounter imaginary numbers in the context of real numbers, or perhaps when taking a limit at a point ...
When dealing with a sum, you have a sequence that generates the terms. In this case, you have the sequence. #a_n = (3/2)^n#
In short, the limit does not exist if there is a lack of continuity in the neighbourhood about the value of interest.
May 9, 2016 · The function, as given, is not continuous at 0 as 0sin(1/0) is not defined. However, we may make a slight modification to make the function continuous, defining f(x) as f(x) = {(xsin(1/x)" if "x!=0),(0" if "x=0):} We will proceed using this modified function. Using the epsilon-delta definition of a limit, we must show that for any epsilon > 0 there exists a delta > 0 such that if 0 < |x - 0| < delta then |f(x) - f(0)| < epsilon To do so, we first let epsilon > 0 be arbitrary. Next, let delta ...
Jun 7, 2017 · By definition a real series {a_n} is bounded if we can find an M in RR and an integer N for which: n > N => abs(a_n) < M For a_n = 2^n first we not that all terms are positive, so that: abs(a_n) = a_n Then we can see that for any M in RR if we choose N > lnM xx ln2 then for n > N: a_n = 2^n > 2^N > 2^(lnM xx ln2) = (2^ln2)^lnM = e^lnM = M Hence for any M in RR we can find a term of the series such that: abs(a_n) > M which proves the series is unbounded.
