Finding Discontinuous Points and Limits for f(x) = (x^2 - 4)/(x - 2)

Click For Summary

Homework Help Overview

The discussion revolves around identifying points of discontinuity for the function f(x) = (x^2 - 4)/(x - 2) and determining the limits as x approaches those points. Participants are exploring concepts related to continuity and limits in the context of rational functions.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the process of factoring the function and the implications of simplifying it. Questions arise regarding the nature of discontinuities, particularly at x = 2, and the reasoning behind identifying multiple points of discontinuity.

Discussion Status

The discussion is active, with participants offering insights into the conditions for continuity and the significance of undefined points. Some guidance has been provided regarding the interpretation of limits and the definition of continuity, though confusion remains about applying these concepts to different functions.

Contextual Notes

There is a noted confusion regarding the rules of simplification and continuity, particularly in relation to the function's behavior at x = 2 and other potential discontinuities in different functions being considered.

staples82
Messages
17
Reaction score
0

Homework Statement


Find all values x=a where the function is discontinuous. For each value of x, give the limit of the function as x approaches a.

f(x)=[x^2-4]/[x-2]

Homework Equations





The Attempt at a Solution


ok the first thing I did was factored out: (x+2)(x-2)/(x-2), then I crossed out the x-2 and ended up with f(x)=x+2...then I checked the answer and the answer was 2,4 I don't understand how they got 2,4 besides that f(x) can't equal 2, because then it'd be undefined.
 
Physics news on Phys.org
When x = 2, function is undefined. (I always confuse continuity concepts, but I think you are not allowed to factor or simplify things)

and limit of function as it approaches x = 2 is 4

Two things that question asked for.
 
Two functions are equal iff they have the same domain, range, and for every x in the domain f(x) = g(x). Thus, if f(x) is your original function, and g(x) = x+2 then f(x) and g(x) are only equal if you restrict g(x) to all real numbers except 2 which is line x+2 with a gap at the point (2,4).
 
I'm still not grasping the concept, I don't understand where we get 2 from besides when 2-2=0, and you can't divide any number into zero.

Also, I'm trying to work other problems, but apparently, it doesn't make sense, such as

[5+x]/[x(x-2)], so if I can't change anything as rootX says, how am I supposed to know my next step?
 
for second: discontinous when x(x-2) = 0

so either x = 0 or x-2 = 0

for first: without changing anything plug in 2, and you get 0/0 (so discontinouty)
 
A function f(x) is continuous at x=a if i) f(a) exists and ii) the limit x->a f(x) exists and iii) the limit equals f(a). Your first example is one where the limit exists (here you can cancel things), but f(a) does not exist. The second is one where the limit doesn't exist and neither does f(a). Does that help? And your second one isn't a 0/0 case, that's the first one.
 
staples82 said:

Homework Statement


Find all values x=a where the function is discontinuous. For each value of x, give the limit of the function as x approaches a.

f(x)=[x^2-4]/[x-2]

Homework Equations





The Attempt at a Solution


ok the first thing I did was factored out: (x+2)(x-2)/(x-2), then I crossed out the x-2 and ended up with f(x)=x+2...then I checked the answer and the answer was 2,4 I don't understand how they got 2,4 besides that f(x) can't equal 2, because then it'd be undefined.
Then go back and read the question again! The question asks for a value of x for which the function is not continuous. It is not continuous at x= 2 precisely because it is undefined there.
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
4
Views
4K
  • · Replies 26 ·
Replies
26
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K