f(x) = x^(2/3) Prove that it is continuous everywhere on [0,1]
Show that, either: \[ \lim_{x\to c^+}x^{\frac{2}{3}}=\lim_{x\to c^-}x^{\frac{2}{3}}=c^{\frac{2}{3}} \]For some \(c\in [0,1]\) Or: \[ \lim_{h\to 0^+}(x+h)^{\frac{2}{3}}=\lim_{h\to 0^+}(x-h)^{\frac{2}{3}}=x^\frac{2}{3} \]\(\forall x \in \mathbb{R}\)
Can I say that Lim x--> c+ f(x) = c^(2/3) Lim x--> c- f(x) = c^(2/3) So Lim x--> c f(x) = c^(2/3) which exists -----> but can i say it is finite?
Yes, and, yes, but I don't know what you mean by 'it is finite.'
As in c^(2/3) is not + or - infinity
Well, you can prove that, if \(a\ge 0\) \[ a^3\ge a^2 \]Which implies: \[ a\ge a^\frac{2}{3} \]Clearly, \(a\) is non-infinite for finite values, hence, since \(a^\frac{2}{3}\) is smaller than \(a\) for all positive real values, it is non-infinite.
Using the lim h--> 0 f(x) way, can you show me how to do Lim h--> 0+ f(x) = Lim h--> 0- f(x)?
Actually, I think for the 2nd way, shouldn't it be Lim h--> 0+( f(x+h) - f(x))/h?
Nevermind! I figured it out
Sure: \[ (x+h)^\frac{2}{3}=\sqrt[3]{x^2+2xh+h^2} \]If we take the limit as \(h\to 0^\pm\): \[ \sqrt[3]{x^2+2xh+h^2}=\sqrt[3]{x^2+2x\cdot0+0^2}=\sqrt[3]{x^2}=x^\frac{2}{3} \]
No, that last part is a derivative, not a definition of continuity.
Wait or was that for derivatives?
In response to: "Actually, I think for the 2nd way, shouldn't it be Lim h--> 0+( f(x+h) - f(x))/h?"
If I'm doing limits, you were right, it should have been lim of f(x) not the slope?
That is correct.
What's the diff between the 2 ways?
They're exactly equivalent. Sometimes, it happens that one is easier to use than the other, but they're no different, mathematically.
Well, I'm off, hopefully that helped.
Thank you!
Quick question: the first way is general and the 2nd way is for a specific point?
I mean the 1st way is specific point and 2nd way is in general?
Yes.
Ok! Thanks!
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