Prove continuity by first principle

AI Thread Summary
To prove that f(x) = x^2 is continuous at x = 2 using the ε - ∂ definition, the discussion focuses on establishing the relationship between |x - 2| and ε through |x + 2|. The key step involves bounding |x + 2| by restricting |x - 2| to be less than 1, which leads to the conclusion that 3 < |x + 2| < 5. This allows for the derivation of the inequality |x - 2| < ε/|x + 2|. Ultimately, it is determined that setting δ = ε/5 ensures the continuity condition is satisfied. The collaborative effort in the discussion highlights the importance of careful bounding in continuity proofs.
chocolatefrog
Messages
11
Reaction score
0

Homework Statement



Prove that f(x) = x^2 is continuous at x = 2 using the ε - ∂ definition of continuity.

2. The attempt at a solution

Using the definition of continuity, I've reached thus far in the question:

|x - 2||x + 2| < ε whenever |x - 2| < ∂

3. Relevant equations

I have no clue how to move forward from here. I know while solving this type of questions, we try to solve the first inequality so that ∂ can be written in terms of ε, but I can't seem to figure out what to do with |x + 2|.
 
Physics news on Phys.org
Okay, |x-2||x+2|&lt; \epsilon and so |x-2|&lt; \epsilon/|x+ 2|. That is almost what you want- you just need to replace that |x+2| with a constant. To do that, take some reasonable restriction on |x- 2|.

We want to take the limit as x goes to 2 so we really are only interested, say, in |x- 2|< 1 ("1" was chosen pretty much arbitrarily). That is the same as saying that -1<x- 2< 1 and so 3< x+ 2< 5. Since 3 itself is larger than 0, that is 3< |x+2|< 5. 1/5< 1/|x+2|< 1/3 so \epsilon/5&lt; \epsilon/|x+2|&lt;\epsilon/3. Now, what must your \delta be to guarantee that "if |x- 2|&lt; \delta, then |x- 2|&lt; \epsilon/|x+2|&amp;quot;?
 
Then \delta = \epsilon/5.
 
Oh, and I totally forgot, thanks a lot for your help! :)
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top