epsilon-delta definition

**Deadstar** · November 11th 2007, 12:05 PM

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

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

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

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

any help appreciated.

**Jhevon** · November 11th 2007, 12:26 PM

Originally Posted by Deadstar

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

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

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

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

any help appreciated.

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

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

now continue

**Deadstar** · November 12th 2007, 03:59 AM

k...

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

$|u^3 - v^3| = |(u - v)(u^2 + uv + v^2)| \le |(u - v)(u^2 + 2uv + v^2)| =$ $|(u - v)|(u + v)^2 < |u - v|(2|y_0| + 1)^2$

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

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.

**Jhevon** · November 12th 2007, 09:45 AM

Originally Posted by Deadstar

k...

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

$|u^3 - v^3| = |(u - v)(u^2 + uv + v^2)| \le |(u - v)(u^2 + 2uv + v^2)| =$ $|(u - v)|(u + v)^2 < |u - v|(2|y_0| + 1)^2$

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

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:

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

i think you can finish

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

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