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

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SUMMARY

The continuity of the natural logarithm function, ln x, over the interval (0, ∞) can be established by defining ln x as the inverse of the exponential function, e^y = x. The proof utilizes the property that ln(x/u) = ln x - ln u, and as u approaches x, the ratio x/u approaches 1, leading to ln(1) = 0. This argument confirms that ln x is continuous for all x > 0, relying on the continuity of the exponential function.

PREREQUISITES
  • Understanding of the natural logarithm function and its properties.
  • Familiarity with the exponential function and its continuity.
  • Basic knowledge of limits and continuity in calculus.
  • Ability to manipulate logarithmic identities.
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  • Study the properties of the exponential function and its inverse, the natural logarithm.
  • Learn about the formal definition of continuity in calculus.
  • Explore proofs of continuity for other functions, such as polynomial and trigonometric functions.
  • Investigate the implications of continuity in real analysis.
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Students of calculus, mathematicians, and anyone interested in understanding the properties of logarithmic functions and their continuity.

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how can we prove that lun x is continuous over (0, \infty )?

Provided that we define : lun x =y <=> e^y =x?
 
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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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