Epsilon-delta definition and continuity at a point

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SUMMARY

The discussion focuses on demonstrating the continuity of the function \( f(y) = \sqrt[3]{y+3} \) at the point \( y = -2 \) using the epsilon-delta definition. Participants clarify that to show continuity, one must prove that for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that \( |f(y) - f(-2)| < \epsilon \) when \( |y + 2| < \delta \). The key steps involve manipulating the inequality to find appropriate bounds for \( y \) in relation to \( \epsilon \). The discussion emphasizes the importance of correctly selecting \( \delta \) based on the derived inequalities.

PREREQUISITES
  • Understanding of the epsilon-delta definition of continuity
  • Familiarity with limits and their properties
  • Basic knowledge of cube root functions
  • Proficiency in manipulating inequalities
NEXT STEPS
  • Study the epsilon-delta definition of continuity in depth
  • Learn how to manipulate inequalities to find bounds in limit proofs
  • Explore examples of continuity proofs for different types of functions
  • Practice problems involving limits and continuity using the epsilon-delta approach
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Students in calculus, mathematics educators, and anyone interested in understanding the formal definition of continuity and its applications in real analysis.

jameswill1am
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Homework Statement



Show that the following equation is continuous using the epsilon-delta definition at y=-2

Homework Equations



[tex]\f(y)=\sqrt[3]{y+3}[/tex]

The Attempt at a Solution



so i got to a stage where;

[tex]\frac{1}{\left|\left(\sqrt[3]{y+3}\right)^{2}+\sqrt[3]{y+3}+1\right|}\times\left|y-c\right|[/tex]

but this is where i always get stuck on these things. I just don't know where to start with picking my delta. If anyone could explain the best way of going about it that would be very helpful.

Thanks

p.s. i didn't type out the whole extent of my solution so far. Hopefully its ok so far but let me know if that bit is wrong as well.
 
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I revised your LaTeX so that it would render as you wanted it to.
jameswill1am said:

Homework Statement



Show that the following equation is continuous using the epsilon-delta definition at y=-2

Homework Equations



[tex]f(y)=\sqrt[3]{y+3}[/tex]

The Attempt at a Solution



so i got to a stage where;

[tex]\frac{1}{\left|\left(\sqrt[3]{y+3}\right)^{2}+\sqrt[3]{y+3}+1\right|}\times\left|y-c\right|[/tex]

but this is where i always get stuck on these things. I just don't know where to start with picking my delta. If anyone could explain the best way of going about it that would be very helpful.

Thanks

p.s. i didn't type out the whole extent of my solution so far. Hopefully its ok so far but let me know if that bit is wrong as well.

I think I understand what you're doing, with the idea being that (a - b)(a2 + ab + b2) = a3 - b3. I tried a different approach that I think will work.

I'm assuming you want to show that [tex]\lim_{y \rightarrow -2} f(y) - f(-2) = 0[/tex]

This means you want to show that for any [itex]\epsilon[/itex] > 0 there is a number [itex]\delta[/itex] > 0 such that |f(y) - f(-2)| < [itex]\epsilon[/itex] when |x - (-2)| < [itex]\delta[/itex]
So given such an [itex]\epsilon[/itex], we want
[tex]|(y + 3)^{1/3} - 1| < \epsilon[/tex]
[tex]-\epsilon + 1 < (y + 3)^{1/3} < \epsilon + 1[/tex]
[tex](-\epsilon + 1)^3 < y + 3 < (\epsilon + 1)^3[/tex]
[tex](-\epsilon + 1)^3 - 1 < y + 2 < (\epsilon + 1)^3 -1[/tex]

All that's left is to figure out what you need to use for your [itex]\delta[/itex].
 

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