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## Homework Statement

Given a sequence [itex](x_n), x_n > 0[/itex] for every [itex]n\in\mathbb{N}[/itex] and [itex]\lim\limits_{n\to\infty} x_n = L > 0[/itex], show that [itex]\ln x_n\to \ln L[/itex] when [itex]n\to\infty[/itex].

## Homework Equations

## The Attempt at a Solution

As logarithm function is an elementary function, meaning it is continuous in its domain [itex]D:= (0,\infty)[/itex] then we have that:

[itex]\forall\varepsilon >0, \exists\delta >0:\forall x\in D\left ( 0<|x-L|<\delta\Rightarrow |\ln x - \ln L|<\varepsilon\right )[/itex]

Provided that [itex]x_n \to L[/itex], then there exists [itex]N\in\mathbb{N}[/itex] such that:

[itex]n\geq N\Rightarrow |x_n-L|<\delta[/itex], from which it follows that:

[itex](n\geq N\Rightarrow |\ln x_n -\ln L|<\varepsilon) \Leftrightarrow \lim\limits_{n\to\infty} \ln x_n =\ln L[/itex]

I googled this problem, but I couldn't find an epsilon-delta argument, so I gave it a go. Is this convincing?

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