Proving a limit to infinity using epsilon-delta

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    Infinity Limit
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Discussion Overview

The discussion revolves around proving the limit of the function 2x + 3 as x approaches infinity using the epsilon-delta definition of limits. The focus is on the theoretical framework and the application of the definition in this context.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant presents the limit statement and seeks guidance on applying the epsilon-delta definition.
  • Another participant requests clarification on the definition of the limit and what needs to be proven.
  • A third participant outlines the formal definition of the limit approaching infinity, specifying the conditions involving d and k.
  • A later reply suggests that the initial value a can be disregarded and provides a method to find k in relation to d, indicating that k can be chosen as (d-3)/2.

Areas of Agreement / Disagreement

Participants are engaged in a constructive dialogue, with no explicit consensus reached yet. Different aspects of the epsilon-delta definition are being explored, but the discussion remains open-ended.

Contextual Notes

The discussion does not fully resolve the application of the epsilon-delta definition, as it relies on specific assumptions about the values of d and k, which may not be universally applicable without further clarification.

brooklysuse
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lim 2x + 3 = ∞.
x→∞

Pretty intuitive when considering the graph of the function. But how would I show this using the epsilon-delta definition?Thanks!
 
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Please expand the definition of the limit and write what you need to prove.
 
Looking to use this definition. f:A->R, A is a subset of R, (a, infinity) is a subset of A.

lim f(x) =infinity if for any d in R, there exists a k>a such that when x>k, then f(x)>d.
x->infinity
 
You may forget about $a$. So you have to prove that for every $d$ there exists a $k$ such that if $x>k$, then $f(x)=2x+3>d$. So consider an arbitrary $d$. You need to show that there exists a $k$ such that $x>k$ implies $x>(d-3)/2$. It's sufficient to take $k=(d-3)/2$.
 

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