Limits and Continuity?

Question

lim
x->a

x^2 + x + 4
-----------------
x^3 - 2x^2 + 7

For values of a does the above function exist? What does this say about the
continuity of this rational function?

To reply to Jim there, this isn't my homework. It's an extra optional assignment, and I personally would like to understand how to solve it.

I realize that to find where a exists, I need to find where it doesn't exist first, which would be where the denomiator is equal to zero, but I don't know how to get that.

Answer 1

Replace x with a.
(a^2 + a + 4) / (a^3 - 2a^2 + 7)
The numerator is always positive.
This limit will not exist when the denominator equals zero.
a^3 - 2a^2 + 7 = 0 when a = -1.4288
The function is discontinuous at x = -1.4288

Answer 2

i'm going to offer you hand-waving (experience-making) solutions fairly than something technical. a million) 2 factors. x=a million and x=2. with the aid of fact that is truthfully the cost of a parabola with x-intercepts at a million and a couple of, fairly than "dip under the axis" at those factors the parabola "bounces" back interior the effective. those 2 "leap factors" are the standards the place f(x) isn't differentiable - they do no longer look to be "mushy." 2) fake. unquestionably the decrease is comparable to a million. logx provides you with what means of 10 the extensive type is and the a million/x exponent will shrink that all the way down to a million. Take a brilliant extensive type as an occasion: one thousand million (a million billion). log one thousand million = 9. Now 9^(a million/one thousand million) is very close to to 9^0 that's a million. in certainty as you develop x, the exponent will become so tiny as to in certainty be 0, on an identical time as the log x section grows lots extra slowly. 3) fake. cos(0) = a million. a million^infinity = a million. Your decrease = a million. i be attentive to that may no longer precisely rigorous therapy of the undertaking, even though it shows the assumption. for terribly super exponents n, the cost of cos(x)^n would be equivalent to 0 different than very close to to x=0. no count how super n is, as you attitude an arbitrarily small distance from x=0, the function will shoot back as much as a million, considering that cos(0)=a million. 4) It effective is. For all numbers between 0 and a million, the two the 1st term and 2nd term around all the way down to 0. At x=a million, the two words equivalent a million, and a million-a million=0. For numbers as we talk after a million, the 1st term rounds all the way down to a million^2, that's a million. the 2nd term may well be, as an occasion, a million.a million^2, that's a million.21, which rounds all the way down to a million (nonetheless 0). that is not till x=sqrt(2) that the 2nd term rounds to 2. for this reason non-quit at x=a million.

Answer 3

You are correct about finding where the denominator is zero. The denominator is x^3 - 2x^2 + 7. This has one real root at x=-1.429.

Answer 4

to find discontinuity of a rat. fxn set denominator = 0 solve for x, these are the values which cause discontinuity simple as that !!

caveat: you must be able to factor the den. since its is a polynomial hopefully u have these skills...

Answer 5

Do your own damn homework. Open that math book and look it up. You'll save more time doing it yourself than posting that crap here.

Answer 6

sorry.i will tryt