How to use epsilon K proof to show a limit using Calculus?

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SUMMARY

The discussion focuses on using the epsilon-K proof to demonstrate that the limit of the function \(\frac{n^2 + 2n + 1}{2n^2 + 3n + 2}\) approaches \(\frac{1}{2}\) as \(n\) approaches infinity. The key inequality established is \(\left| \frac{n^2 + 2n + 1}{2n^2 + 3n + 2}-\frac{1}{2}\right| \leq \frac{1}{2n}\), which is crucial for the proof. A clarification is provided that if \(\frac{1}{2n} < \epsilon\), then it follows that \(\frac{1}{2(2n+1)} < \epsilon\), reinforcing the validity of the limit.

PREREQUISITES
  • Understanding of epsilon-delta definitions in calculus
  • Familiarity with limits and continuity concepts
  • Basic algebraic manipulation skills
  • Knowledge of sequences and their convergence
NEXT STEPS
  • Study the epsilon-delta definition of limits in calculus
  • Explore examples of epsilon-K proofs for different functions
  • Learn about the properties of limits involving rational functions
  • Practice solving limit problems using algebraic techniques
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Students studying calculus, particularly those focusing on limits and proofs, as well as educators looking for examples of epsilon-K proofs in teaching materials.

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Homework Statement


Use [tex]\epsilon[/tex] K proof to show:
[tex]lim \left(\frac{n^2 + 2n + 1}{2n^2 + 3n + 2}\right) = \frac{1}{2}[/tex]


Homework Equations


Hint first show
[tex]\left| \frac{n^2 + 2n + 1}{2n^2 + 3n + 2}-\frac{1}{2}\right| \leq \frac{1}{2n}, \hspace{0.5cm} n\epsilon N[/tex]



The Attempt at a Solution


See the pdf. Please let me know if my argument can be more thorough.
 

Attachments

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At one point you have
[tex]\frac{1}{n(2n+1)}< \epsilon \doublearrow \frac{1}{2n}< \epsilon[/tex]
What you should say is "Because
[tex]\frac{1}{2(2n+1)}< \frac{1}{2n}[/tex]
if
[tex]\frac{1}{2n}< \epsilon[/tex]
it will be true that
[tex]\frac{1}{2(2n+1)}< \epsilon[/tex]
 

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