# Defintion of a limit by Serge Lang - unusual??!!

• Apr 17th 2010, 10:44 PM
Bernhard
Defintion of a limit of a function by Serge Lang - unusual??!!
I am working through Serge Lang's book "Undergraduate Analysis" (Second Edition)

On pagfe 42 he defines the limit of a function as follows:

"We shall say that the limit of f(x) as x approaches a exists if there exists a number L having the following property. Given $\epsilon$, there exists a number $\delta$ > 0 such that for all x $\in$ S satisfying

|x - a| < $\delta$

we have

|f(x) - L| < $\epsilon$ {see below for some
obvious and immediate consequences}

This definition seems at odds with the definition in Apostal and other texts snice they do not allow x to actually assume the value a and write something equivalent to

0 < |x - a| < $\delta$

Am I right in assuming Lang's definition differs?

If so - are there significant consequences for theorems - I mean does one constantly have to be careful over this matter?

{Note further that on page 43 Lang writes:

"Next, suppose a is an element of S. We consider any function f on S.

Then the limit $lim x_{\rightarrow a} f(x)$ exists.

We contend that it must be equal to f(a)"

Surely (as a consequence of Lang's defn) this is not the usual conclusion!

Yet further ... on page 44 Lang writes:

"Define g on S by g(x) = x if x ne 0 and g(0) = 1. Then $lim_{x \rightarrow 0} f(x)$ does not exist. Again with the defn of Apostal and others $lim_{x \rightarrow 0} f(x)$ would be equal to 0 even though the value of the function at 0 is 1."

Is my reasoning correct?

Is it right to be alarmed at the possible consequences of this different definition for being able to follow mainstream analysis? Should I switch to another text? }

Bernhard
• Apr 17th 2010, 11:24 PM
CaptainBlack
Quote:

Originally Posted by Bernhard
I am working through Serge Lang's book "Undergraduate Analysis" (Second Edition)

On pagfe 42 he defines the limit of a function as follows:

"We shall say that the limit of f(x) as x approaches a exists if there exists a number L having the following property. Given $\epsilon$, there exists a number $\delta$ > 0 such that for all x $\in$ S satisfying

|x - a| < $\delta$

we have

|f(x) - L| < $\epsilon$

This has an obvious problem as it appears to require that $f(a)$ be defined, so this definition does not allow us to conclude that:

$\lim_{x \to 0} \frac{\sin(x)}{x}=1$

That is unless $S$ is defined to exclude $a$ in this case, but you will note that Serge requires $f$ to be defined on $S$ and $a$ adherent to $S$ (which is essentially that $a$ is a limit point of $S$). So here all is OK.

However there is still a problem:

Define:

$\text{sinc}(x)=\begin{cases}\frac{\sin(x)}{x},&x \ne 0 \\ 7, & x=0 \end{cases}$

With $S=\mathbb{R}$

Now Serge's definition of a limit means $\lim_{x \to 0}\text{sinc}(x)$ does not exist, while the traditional definition has $\lim_{x \to 0}\text{sinc}(x)=1\ne \text{sinc}(0)$. This is essentially what he is pointing out on page 44 (and at the bottom of that page he does also point out that his convention is different from some other authors - I think I would say than most others myself).

CB
• Apr 17th 2010, 11:51 PM
Bernhard
Defintion of a limit of a function by Serge Lang - unusual??!!
Thanks CB

I presume that, under Serge Lang's definition, if the Set S includes 0 and $\frac{sine\ x}{x}$ is defined as 1 at 0 then $lim_{x \rightarrow 0}\frac{sine\ x}{x}$ = 1. If $\frac{sine\ x}{x}$ is defined as something else at 0 then according to Lang the limit does not exist.

On the other hand if the set S does not include 0 [but includes all points around 0] then the limit exists and is equal to 1.

Does that sound right?
• Apr 18th 2010, 12:18 AM
CaptainBlack
Quote:

Originally Posted by Bernhard
Thanks CB

I presume that, under Serge Lang's definition, if the Set S includes 0 and $\frac{sine\ x}{x}$ is defined as 1 at 0 then $lim_{x \rightarrow 0}\frac{sine\ x}{x}$ = 1. If $\frac{sine\ x}{x}$ is defined as something else at 0 then according to Lang the limit does not exist.

On the other hand if the set S does not include 0 [but includes all points around 0] then the limit exists and is equal to 1.

Does that sound right?

Sound right (see his discussion of this in the pages immediately following p44)

CB