How can we prove the continuity of ln x over (0, ∞)?

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Discussion Overview

The discussion centers on proving the continuity of the natural logarithm function, ln x, over the interval (0, ∞). Participants explore various approaches and assumptions related to this proof, including the relationship between ln x and the exponential function.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant suggests defining ln x in terms of the exponential function, proposing that if ln x = y, then e^y = x.
  • Another participant presents a relationship, ln x - ln u = ln(x/u), and discusses the limit as u approaches x, concluding that this leads to ln(1) = 0.
  • A later reply requests further elaboration on the proof, indicating uncertainty about how to begin the argument.
  • One participant expresses uncertainty about the initial assumptions, questioning whether the continuity of the exponential function is being assumed and noting a potential flaw in their own proof that relies on the continuity of ln(x) at x = 1.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the proof or the assumptions required for continuity. Multiple viewpoints and approaches are presented, indicating ongoing debate and exploration of the topic.

Contextual Notes

Some assumptions regarding the continuity of related functions, such as the exponential function, are not clearly established. The discussion also reflects varying levels of confidence in the proposed proofs and their completeness.

Who May Find This Useful

Readers interested in mathematical proofs, particularly in the context of calculus and the properties of logarithmic functions, may find this discussion relevant.

evagelos
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how can we prove that lun x is continuous over (0, [tex]\infty[/tex] )?

Provided that we define : lun x =y <=> [tex]e^y =x[/tex]?
 
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lnx - lnu = ln(x/u). For any fixed x > 0, u->x => x/u -> 1 and ln1=0.

You can dress this proof up.
 


mathman said:
lnx - lnu = ln(x/u). For any fixed x > 0, u->x => x/u -> 1 and ln1=0.

You can dress this proof up.

Can you elaborate a little more ? I do not how to start
 


I am not sure what you are given to start with. For example are you assuming ey is continuous? My proof (I admit) is somewhat flawed. It needs continuity of ln(x) for x=1, which then implies continuity for all x.
 

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