Proof: sqrt(x) is Continuous using Epsilon Delta Definition | Real Analysis Exercises

We prove that f(x)=sqrt(x), the square root function, is continuous on its entire domain where it is real, from 0 to infinity including 0. We complete this proof using the epsilon delta definition of continuity of a function at a point. To do this, we simply take an epsilon greater than 0 and an arbitrary point c from our domain, then go through the motions of finding a delta greater than 0 so that any x in D that is within delta of c has an image within epsilon of c's image. However, our approach will actually force c into the denominator in our proof, so we will need to take care of continuity at 0 separately. Let me know if there are more epsilon delta continuity proofs you want to see! #RealAnalysis

#RealAnalysis #Math

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[Chilliak]

you are 30x better than every upper division college professor ive had, and they get paid thousands. American education system !!!

[ChaloGhat]

If x and c are smaller than 1 then the inequality at 7:24 holds ?

[peteryan]

Hi, thanks for the video. For the limit at 0, weren't we supposed to prove only the right hand limit, since the left hand limit doesn't exist?

[machitoons]

the main thing I always found confusing about εδ proofs is how "backwards" they feel, in the sense you have to be veeery careful about what things you assume in what order, and it's easy to make mistakes. I'm leaving a way I like to do it, in case others find it easier too:
I find doing εδ proofs easier to do via contraposition, i.e. we assume 0 < ε ≤ |f(x)-f(c)| instead and compute a suitable δ directly. (|f(x) - f(c)|≥ε ⇒ |x-c|≥δ) ⇔ (|x-c|<δ ⇒ |f(x)-f(c)|<ε)

example for cont. of √-function:
let D = [0, ∞)

let c = 0,
let ε > 0, let x ∈ D, assume ε ≤ |√x - √0| ∴
0 < ε ≤ |√x - √0| = √x ⇒ ε² ≤ (√x)² = x = |x - 0|  , let δ = ε² then δ ≤ |x - 0|

⇒ ( ε ≤ |√x - √0| ⇒ δ ≤ |x - 0| ) ⇔ ( |x - 0| < δ ⇒ |√x - √0| < ε ),  ergo √-function is cont. at 0

let c > 0 (the rest of D)
let ε > 0, let x ∈ D, assume ε ≤ |√x - √0| ∴
0 < ε ≤ |√x - √c| = |√x - √c|
let's multiply by |√x + √c| ≥ 0:
ε|√x + √c| ≤ |√x - √c|·|√x + √c| = |√x² - √xc + √xc - √c²| = |x - c|
ε(√x + √c) ≤ |x - c|,  but since √x and √c are both positive, we know ε√c ≤ ε√x + ε√c, ergo ε√c ≤ |x-c|
let δ = ε√c, then via contraposition, again:
... ( |x - c| < δ ⇒ |√x - √c| < ε )  ☐

I mean it's the same steps, with the only difference being that we don't need to work with *strict* inequalities (which makes it easier, imo) as much (besides making sure the δ we get is infact >0) and that we can compute δ directly instead of needing to assume |x-a| to be strictly less than a unknown δ, it's a tiny bit less mentally taxing I think, yet I never see εδ-proofs done via contrapos., can anyone tell me why?

[icsetricks]

This video is just brilliant.crisp and to the point

[premkumar-soff]

Well this is kind of proof method used in calculus courses. But I think it's real analysis course the proof should be in terms of sequential convergence. But very clear and nice explanation in your videos.

[alpha_inno]

If we apply the epsilon-delta argument considering the domain, is x^1.5 differentiable at x=0?

[rmw]

... perhaps a dumb question, but how do you use the epsilon delta definition to prove that a function is NOT continuous on its domain?

[chair]

Can we prove that it is uniformly continuous with the same method?

[cicerohitzschky]

What is the software, please? Very Nice the video!

[anupamvashishtha]

Please Be My Prof for rest of my degree

[ritebcreative]

well those are some interesting captions...

[odobenusrosmarus]

would it be ok to say delta/sqrtc < delta and then set delta=epsilon?