Converging sequence problem

[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

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