Límites y continuidad

Conceptos y Principios Básicos

The limit concept

¿Is continuous at x=0?¿Does it have any límit?
The limit is one of most important concepts in calculus. Definitions such of derivate and integral are based in this important concept.

Shortly formally, the limit of a real function f in a point a, is the value wich function tends when x tend to the point a. It mains the tendence of function f in a neighborhood of x.

A more rigorous definition is

Limit definition
Lets f a real variable function. It is said that the limit of f at \( x_{0} \) is L when
$$ \forall \epsilon>0, \exists \delta > 0 : |x - x_{0}| < \delta \Rightarrow |f(x) - L | < \epsilon .$$ It is denoted as $$ \lim_{x \mapsto x_{0}} f(x) = L $$


Continuity concept deals with neighborhood. Shortly formally, continuity means that if two ponits x, y are neighbors then its images through f, f(x) and f(y) are also neighbors.

Limit is very related with continuity concept because a function f is continuous at a point \( x_{0} \) following two conditions holds:

   1.-Function f has finite limit L at \( x_{0} \).
   2.- Value of function at \( x_{0} \) is defined and it´s value is L.

Both conditions can be written more fomally as follows:
Continuity definition

Lets f a real variabled function. It is said that f is continuous at \( x_{0} \) if
$$ \forall \epsilon>0, \exists \delta > 0 : |x - x_{0}| < \delta \Rightarrow |f(x) - f(x_{0}) | < \epsilon .$$ Or equivalently
$$ \lim_{x \mapsto x_{0}} f(x) = f(x_{0}) $$

It is obvious the fact that a function f is continuous at \( x_{0} \) then it has limit at this point. Converse does not holds, ie, a funcion can have limit at a certain point, but not neccesarilly be continuous in such point as showed in following example:

Example 1
Lets f(x) defined as follows.

\( f(x) = \left\{\begin{matrix} 1 & x \neq 0 \\ 0 & x = 0 \end{matrix}\right. \)

\( $$ \lim_{x \mapsto 0} f(x) = 1 \)

but \(f(0)=0\)

So, function has limit at x=0 but it is not continuous in such point.

Example 2

Lets f(x) defined as follows.

\( f(x) = \left\{\begin{matrix} \frac{x^2}{2} & x \leq 1 \\ 1-x & x>1 \end{matrix}\right. \)

I've found an interactive graphic for this case (you can drag into):

In this case, function is not continuous and nor has limit at x=1, lets show it:

Note that in this interactive graphic you can drag the x point and also play with the \( \delta \) value. If you drag the point to x=1, for its neighboring points points \( (1-\delta, 1+\delta) \) there not exsits any \(\epsilon\) for wich

\( |x-1|<\delta \Rightarrow |f(x) - f(1)|< \epsilon \).

In other words, \(\delta\) can be as small as we want, but \(\epsilon \) cannot be smaler that jump of function at x=1.

We want to meet other kind of imit definitions:

Infinite limit definitions
Lets f a real variable function. $$ \lim_{x \mapsto \infty }f(x)=L \Leftrightarrow \forall \epsilon> 0, \exists M > 0 : x>M \Rightarrow |f(x) - L | < \epsilon .$$ $$ \lim_{x \mapsto x_{0} }f(x)=\infty \Leftrightarrow \forall M> 0, \exists \delta > 0 : |x-x_{0}|< \delta \Rightarrow f(x) > M .$$ It is analogous to define negative infinite limits
Lateral limits definitions
Lets f a real variabled function.

Rigth limit definition: $$ \lim_{x \mapsto a^{+} }f(x)=L \Leftrightarrow \forall \epsilon> 0, \exists \delta > 0 : x+a <\delta \Rightarrow |f(x) - L | < \epsilon .$$ Left limit definition: $$ \lim_{x \mapsto a^{-} }f(x)=L \Leftrightarrow \forall \epsilon> 0, \exists \delta > 0 : a-x <\delta \Rightarrow |f(x) - L | < \epsilon .$$

There is an important theorem that sometimes it is used as definition of limit. It say that the limit of a function f exists then both lateral limits also exists and are equaly and converse also holds, more formally:

Lets f a real variabled function. If $$ \exists L : \lim_{x \mapsto a }f(x)=L \Leftrightarrow \lim_{x \mapsto a^{-} }f(x)=L=\lim_{x \mapsto a^{+} }f(x) $$

Lets the function \( f(x)=\frac{1}{x^2}\) then

\( \lim_{x \mapsto 0 }\frac{1}{x^2}= \infty \)

Funtion has infinite positive limit at x=0.


\(f(x)=\frac{1}{x}\) then

\( \lim_{x \mapsto 0 }\frac{1}{x}= ? \)

In fact funtion f, has not limit at x=0 because lateral limits does not match: right limit when x maps to 0 is infinite but left limit when x maps to 0 is -infinite.

Note that, in this last two examples, lateral limits matching in first case, but no in second case.

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