Give an example of a function with both a removable and a non-removable discontinuity.?
f(x) = (x-1) /((x-1)(x-2)) 1 is a removable discontinuity ; 2 is a non-removable discontinuity. http://math.stackexchange.com/questions/...
Can you give me examples of non/removable discontinuity at x=2 and x=-2?
a.) limit (for x->2-0)f(x) or limit (for x->2+0)f(x) doesn't exist or equals infinity. for example f(x) = 1/(x-2): limit (for x->2-0)f(x) = - infinity, limit (for x->2+0)f(x) = infinity. b.) There exist both limits: limit (for x->-2-0)f(x)=L1 or limit (for x->-2+0)f(x)=L2 and L1,L2 doesn't equal infinity. Example: f(x) = { x+1 (for x <= -2), x (for x > -2) } then limit (for x->-2-0)f(x) = -1, limit (for x->-2+0)f(x) = -2 c.) Example: f(x) = { x+1 (for x <= -2), 1/(x-2) (for x > -2) }
Give an example of a function with both a removable and a non-removable discontinuity.?
Let's consider the function f(x) = (x + 1)(x + 2) / (x + 2) Now, since there's and x+2 in the denominator, we can't evaluate the function at x = -2. That would make the denominator zero, and we can't divide by zero. There's a discontinuity there. But we can cancel out x + 2 in the numerator and the denominator. So this is a "removable" discontinuity - the function is indistinguishable from g(x) = x+1, except for the discontinuity. Note that, as you get close to x = -2, f(x) doesn't go to infinity. It just approaches the "hole" in the function. A non-removable discontinuity is one that you can't get rid of by canceling. So, h(x) = (x+1) / (x+2) has a non-removable discontinuity at x = -2. As you get close to x = -2, the value of h(x) blows up. I hope that helps! Edit: It occurs to me that maybe you mean a single function that has both a removable and a non-removable discontinuity. f(x) = (x + 2) / [(x + 2)(x + 1)] would work for that - the discontinuity at x = -2 is removable, and the discontinuity at x = -1 is non-removable.
Give an example of a function with at least one removable and one non-removable discontinuity?
both of your limits are examples of functions that have both of these discontinuities. you need to factor the numerator and denominator to help you see. also, graph them on the calculator. the first function factors into (x+3)(x-2) / (x+3)(x-3) This function has a discontinuity at both 3 and -3 as the function is not defined there. however, when you look at the graph you see that there is a vertical asymptote at 3. You can see on the graph, but there is simply a "hole" at -3. We could define this function at -3 to be -5/-6 (just cancel out the (x+3) and plug in -3) and it would fill in the hole and be continuous. So this is a removable discontinuity. However, no matter how we would define the function at 3, there will still be a vertical asymptote. We can't fix that, so this is a non-removable discontinuity. You can make lots of these by just starting in the factored form.
An example of a function with at least one removable and one non-removable discontinuity?
Links that give examples of both removable and unremovable discontinuities: http://answers.yahoo.com/question/index?qid=20080909112908AATC96n http://www.transtutors.com/calculus-homework-help/limits-and-continuity/discontinuities.aspx Good luck!
What is removable discontinuity? How do you identify functions which exhibit such?
That depends on the type of function.Suppose you have two polynomials in a fraction:[math]\displaystyle \frac{p(x)}{q(x)}[/math]The first step would be to factor [math]q(x)[/math].The next step would be to check if these factors also appear in [math]p(x)[/math].If they do you may have found a removable discontinuity.If they don't you found a nonremovable discontinuity.For instance:[math]f(x)=\displaystyle \frac{x^2–1}{x+1}[/math].[math]x+1[/math] has root [math]x=-1[/math].In this case it is removable because the numerator and denominator share the factor [math]x+1[/math].[math]f(x)=\displaystyle \frac{(x-1)(x+1)}{x+1}[/math]If [math]x \to -1[/math], so [math]x \ne -1[/math], we can eliminate this common factor to arrive at:[math]\displaystyle \frac{(x-1)(x+1)}{x+1}=x-1[/math]The value of which is equal to -2 if we substitute [math]x=-1[/math].So if we define:[math]g(x)=x-1[/math]We have removed the singularity at [math]x=-1[/math].You can't remove a discontinuity if a factor occurs more often in the denominator than in the numerator.Now I could type away here and explain about logarithms, goniometric functions and all other types of functions. But the general remark is that one should be aware of the domain. When is a function defined, when do you get into trouble?[math]\sin(x)=0[/math] with [math]x \in [-\pi/2,\pi/2][/math],which [math]x[/math]-value?[math]x/\sin(x)[/math] has which possible singularity on this interval?It helps if you know how to compute limits.
Pre calc: Give an example of a function with at least one removable and one non-removable discontinuity (that?
Let f(x) = (x - 1)/[x(x - 1)]. f has a removable discontinuity at x = 1 (by taking limits, we may "remove" this discontinuity by setting f(1) = 1). Since f has a vertical asymptote at x = 0, this forms a removable discontinuity at x = 0 for f. Otherwise, f is continuous. (Essentially, this is the standard rectangular hyperbola y = 1/x with a removable singularity inserted at x = 1.) I hope that helps!