Proving Continuity at a Point using (ε,δ) Method

In summary, the conversation focuses on a problem involving an (ε,δ) proof of the continuity of f(x) = 1/\sqrt{10 - x^2} at x = -1. The individual has attempted to solve the problem but is unsure how to proceed from the expression ((x^2) - 1)/(3\sqrt{10 - x^2}). They are seeking help from a forum and have received a welcoming response. The solution involves using the standard trick of writing the expression as (x+1) multiplied by the rest, and using ε,δ to minimize (x+1) while keeping the rest bounded.
  • #1
ferret93
1
0
I'm working on a problem as part of exam revision, but I've run into a bit of trouble so far. The problem is;
Give an (ε,δ) proof that f(x) = 1/[itex]\sqrt{10 - x^2}[/itex] is continuous at x = -1

The attempt at a solution
So far what I've gotten is f(x) - f(-1) = 1/([itex]\sqrt{10 - x^2}[/itex]) - 1/3
= (3 - ([itex]\sqrt{10 - x^2}[/itex]))/(3[itex]\sqrt{10 - x^2}[/itex])
= ((x^2) - 1)/(3[itex]\sqrt{10 - x^2}[/itex])

Then from here I've gotten -2 < x < 0 → 10 - x^2 > 6
i'm lost from where to go from here though i just can't see a way though, any help will be greatly appreciated.
 
Last edited:
Physics news on Phys.org
  • #2
welcome to pf!

hi ferret93! welcome to pf! :smile:
ferret93 said:
= ((x^2) - 1)/(3[itex]\sqrt{10 - x^2}[/itex])

the standard trick is to write that as (x+1) times the rest …

you use ε,δ to minimise (x+1), and some other limit to kepp the rest bounded :wink:
 

Question 1: What is a "continuity at a point" problem?

A continuity at a point problem is a mathematical concept that focuses on the behavior of a function at a specific point. It involves determining if a function is continuous at a given point, which means that the function's value at that point is equal to the limit of the function as the independent variable approaches that point.

Question 2: How can I determine if a function is continuous at a point?

To determine if a function is continuous at a point, you need to check three conditions: the function must be defined at the point, the limit of the function as the independent variable approaches the point must exist, and the function's value at the point must be equal to the limit. If all three conditions are met, the function is considered continuous at that point.

Question 3: What is the importance of continuity at a point in mathematics?

Continuity at a point is an important concept in mathematics because it allows us to understand the behavior of a function at a specific point. It helps us determine the existence of a limit, which is crucial in many mathematical calculations and applications, such as finding derivatives and solving optimization problems.

Question 4: Can a function be continuous at a point but not continuous on an interval?

Yes, a function can be continuous at a point but not continuous on an interval. This means that the function satisfies the three conditions for continuity at the specific point, but it may have discontinuities at other points within the interval. For example, a function with a removable discontinuity at a single point is considered continuous at that point but not on the entire interval.

Question 5: How can I solve a continuity at a point problem?

To solve a continuity at a point problem, you need to follow the three conditions mentioned earlier. First, check if the function is defined at the given point. Then, find the limit of the function as the independent variable approaches that point. Finally, compare the function's value at the point to the limit. If they are equal, the function is continuous at that point; if not, the function is not continuous at that point.

Similar threads

  • Calculus and Beyond Homework Help
Replies
27
Views
726
  • Calculus and Beyond Homework Help
Replies
4
Views
684
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
300
  • Calculus and Beyond Homework Help
Replies
16
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
595
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
839
  • Calculus and Beyond Homework Help
Replies
23
Views
1K
Back
Top