Does Convergence of x_n to L Imply Convergence of ln(x_n) to ln(L)?

[nuuskur]

Homework Statement

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

Homework Equations

The Attempt at a Solution

As logarithm function is an elementary function, meaning it is continuous in its domain D:= (0,\infty) then we have that:
\forall\varepsilon >0, \exists\delta >0:\forall x\in D\left ( 0<|x-L|<\delta\Rightarrow |\ln x - \ln L|<\varepsilon\right )
Provided that x_n \to L, then there exists N\in\mathbb{N} such that:
n\geq N\Rightarrow |x_n-L|<\delta, from which it follows that:
(n\geq N\Rightarrow |\ln x_n -\ln L|<\varepsilon) \Leftrightarrow \lim\limits_{n\to\infty} \ln x_n =\ln L

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

 

[Ray Vickson]

Homework Statement

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

Homework Equations

The Attempt at a Solution

As logarithm function is an elementary function, meaning it is continuous in its domain D:= (0,\infty) then we have that:
\forall\varepsilon >0, \exists\delta >0:\forall x\left ( 0<|x-L|<\delta\Rightarrow |\ln x - \ln L|<\varepsilon\right )
Provided that x_n \to L, then there exists N\in\mathbb{N} such that:
n\geq N\Rightarrow |x_n-L|<\delta, from which it follows that:
(n\geq N\Rightarrow |\ln x_n -\ln L|<\varepsilon) \Leftrightarrow \lim\limits_{n\to\infty} \ln x_n =\ln L

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

Are you allowed to use the fact (as you have done) that $\ln x$ is continuous for $x > 0$, or are you essentially required to prove that first?

 

[nuuskur]

Define f:\mathbb{R}\to\mathbb{R} as f(x) = a^x, a >0. We claim that f(x) is continuous in \mathbb{R}.
Fix z\in\mathbb{R}. If x\to z, then x-z\to 0 and \lim\limits_{x\to z} a^{x-z} = 1 from which:
\lim\limits_{x\to z}a^x = \lim\limits_{x\to z} a^za^{x-z} = a^z\lim\limits_{x\to z}a^{x-z} = a^z.
We have established that f is continuous in (-\infty ,\infty)

Bolzano-Cauchy theorem: Let f be continuous in D\subseteq\mathbb{R}. If y_1, y_2 are two different values of the function, then every y between y_1 and y_2 is also a value of the function.
Therefore the set of values of the function f that is continuous in some interval, is also an interval.

We have a function f: (-\infty, \infty)\to (0,\infty ), which is continous. Its inverse (implies it is invertible - bijection, which it is if a>1 or a<1) is defined to be the logarithm function: f^{-1}: (0, \infty)\to (-\infty,\infty).
f^{-1}: = \log _a (x) is also continuous (f is strictly monotone for a\neq 1, its inverse is also strictly monotone.)

 

[geoffrey159]

Google 'sequential definition of continuity'.

 

[nuuskur]

Google 'sequential definition of continuity'.

English -.- We call the same thing as "Heine's criterion for continuity"

 

[geoffrey159]

As logarithm function is an elementary function, meaning it is continuous in its domain D:= (0,\infty) then we have that:
\forall\varepsilon >0, \exists\delta >0:\forall x\left ( 0<|x-L|<\delta\Rightarrow |\ln x - \ln L|<\varepsilon\right )
Provided that x_n \to L, then there exists N\in\mathbb{N} such that:
n\geq N\Rightarrow |x_n-L|<\delta, from which it follows that:
(n\geq N\Rightarrow |\ln x_n -\ln L|<\varepsilon) \Leftrightarrow \lim\limits_{n\to\infty} \ln x_n =\ln L

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

And yes your proof looks OK to me except that you must mention that $L>0$ and also here

\forall\varepsilon >0, \exists\delta >0:\forall x\in D ...