Advanced Calc/Analysis: Delta Epsilon proof

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SUMMARY

The discussion focuses on proving the limit of the function lim x->0 x^n*sin(1/x) for all natural numbers n using the Delta Epsilon definition of limits. The initial proof for n=1 shows that the limit is zero. The user proposes using mathematical induction to extend the proof to n+1, while seeking guidance on selecting the appropriate limit L and effectively applying the definition of a limit. The discussion emphasizes that bounding the function between two limits can establish the existence of the limit without needing the Squeeze Theorem.

PREREQUISITES
  • Understanding of the Delta Epsilon definition of limits
  • Familiarity with mathematical induction
  • Knowledge of trigonometric functions, specifically sine
  • Basic calculus concepts, including limits and continuity
NEXT STEPS
  • Study the Delta Epsilon definition of limits in depth
  • Practice mathematical induction with various functions
  • Explore the properties of the sine function and its boundedness
  • Learn about limit transformations, such as substituting x=1/t
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Students studying calculus, particularly those tackling limits and proofs, as well as educators seeking to enhance their understanding of mathematical induction and limit definitions.

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Homework Statement


Using the definition of |x-a|<delta implies |f(x) - L|<epsilon, prove that lim x->0 x^n*sin(1/x) holds for all n belonging to natural numbers.

Homework Equations


Definition of a limit

The Attempt at a Solution


Ok, so when I see "prove for all n belonging to natural numbers" I immediately think induction. So this is what I have done so far. For n=1 lim x->0 x^n*sin(1/x) is true; the limit is zero. Now I will assume lim x->0 x^n*sin(1/x) for some n is true, then I need to show that n+1 is also true. So I start using the definition of the limit and don't know what my L should be and how to use induction along with this definition of a limit. Please help and thank you in advanced.
 
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[tex]\mbox{If you can prove that }\lim_{x\to 0}x^{n}sin\left(\frac{1}{x}\right)\mbox{ is bounded by two functions whose limit is L, then you've proven that the limit exists.}[/tex]
[tex]\mbox{(You don't have to invoke the Squeeze Theorem.)}[/tex]

[tex]\mbox{Start with }-1\leq \sin\left(\frac{1}{x}\right)\leq 1\mbox{. What can you say about }x^{n}sin\left(\frac{1}{x}\right)\mbox{ ?}[/tex]

[tex]\mbox{It isn't completely necessary, but it might help you see what L to use if you make the substitution x=1/t, then }[/tex]

[tex]\lim_{x\to 0}x^{n}sin\left(\frac{1}{x}\right) = \lim_{t\to\infty}\left(\frac{1}{t}\right)^{n}sin\left(t\right)[/tex]
 

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