1. ## epsilon-delta definition

I have a question on a tutorial involving using the $\displaystyle \epsilon - \delta$ definition to prove a function is continuous. I missed the lectures on this so im not sure about it...

Let $\displaystyle f(y) = \sqrt[3] {y +3}$ and b > -3

Prove, from the $\displaystyle \epsilon - \delta$ definition, that f(y) is continuous at y = b.
[hint use the identity $\displaystyle u^3 - v^3 = (u - v)(u^2 + uv + v^2)$]

Prove, from the $\displaystyle \epsilon - \delta$ definition, that f(y) is continuous from the right at y = -3

any help appreciated.

I have a question on a tutorial involving using the $\displaystyle \epsilon - \delta$ definition to prove a function is continuous. I missed the lectures on this so im not sure about it...

Let $\displaystyle f(y) = \sqrt[3] {y +3}$ and b > -3

Prove, from the $\displaystyle \epsilon - \delta$ definition, that f(y) is continuous at y = b.
[hint use the identity $\displaystyle u^3 - v^3 = (u - v)(u^2 + uv + v^2)$]

Prove, from the $\displaystyle \epsilon - \delta$ definition, that f(y) is continuous from the right at y = -3

any help appreciated.
You want to show that For all $\displaystyle \epsilon > 0$ there exists a $\displaystyle \delta > 0$ such that $\displaystyle y \in dom(f)$ and $\displaystyle |y - b|< \delta$ implies $\displaystyle |f(y) - f(b)| < \epsilon$

here, use $\displaystyle u = \sqrt[3]{y + 3}$ and $\displaystyle v = \sqrt[3]{b + 3}$

now continue

3. k...

Let $\displaystyle \epsilon > 0$, choose some $\displaystyle \delta > 0$ to be determines such that $\displaystyle |y - y_0| < \delta$ when $\displaystyle y \in dom(f)$. Upper bound for $\displaystyle |y + y_0|$ given that $\displaystyle |y - y_0| < \delta$ so let us say $\displaystyle |y - y_0| < 1 => |y| < |y_0 + 1|$. So by triangle inequality, $\displaystyle |y + y_0| \le |y| + |y_0| < 2|y_0| + 1$. Now consider $\displaystyle |f(y) - f(y_0)|$. We have:

$\displaystyle |u^3 - v^3| = |(u - v)(u^2 + uv + v^2)| \le |(u - v)(u^2 + 2uv + v^2)| = $$\displaystyle |(u - v)|(u + v)^2 < |u - v|(2|y_0| + 1)^2 \displaystyle \delta = min{ 1,\displaystyle epsilon/\displaystyle (2|y_0| + 1)^2} sorry but im not that great at the math equation thing yet so the last bit is a bit muddled. And somehow my font has gone red... weird... After this (if its right) i dont really know how to end it. 4. Originally Posted by Deadstar k... Let \displaystyle \epsilon > 0, choose some \displaystyle \delta > 0 to be determines such that \displaystyle |y - y_0| < \delta when \displaystyle y \in dom(f) . Upper bound for \displaystyle |y + y_0| given that \displaystyle |y - y_0| < \delta so let us say \displaystyle |y - y_0| < 1 => |y| < |y_0 + 1|. So by triangle inequality, \displaystyle |y + y_0| \le |y| + |y_0| < 2|y_0| + 1. Now consider \displaystyle |f(y) - f(y_0)|. We have: \displaystyle |u^3 - v^3| = |(u - v)(u^2 + uv + v^2)| \le |(u - v)(u^2 + 2uv + v^2)| =$$\displaystyle |(u - v)|(u + v)^2 < |u - v|(2|y_0| + 1)^2$

$\displaystyle \delta = min${ 1,$\displaystyle epsilon$/$\displaystyle (2|y_0| + 1)^2$}

sorry but im not that great at the math equation thing yet so the last bit is a bit muddled. And somehow my font has gone red... weird...

After this (if its right) i dont really know how to end it.
i don't think an upper bound like that is needed here. and you started off your list of inequalities wrong. |f(y) - f(b)| is not |u^3 - v^3|

here's how you would begin:

$\displaystyle |f(y) - f(b)| = |\sqrt[3]{y + 3} - \sqrt[3]{b + 3}| \le$ $\displaystyle \left| (\sqrt[3]{y + 3} - \sqrt[3]{b + 3}) \left( (y + 3)^{\frac 23} + \sqrt[3]{(y + 3)(b + 3)} + (b + 3)^{\frac 23} \right) \right|$ $\displaystyle = |(y + 3) - (b + 3)| = |y - b|$

i think you can finish

i guess you would also need to show that $\displaystyle \left( (y + 3)^{\frac 23} + \sqrt[3]{(y + 3)(b + 3)} + (b + 3)^{\frac 23} \right) \ge 1$