**The limit concept**

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

¿Is continuous at x=0?¿Does it have any límit?

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**

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:

**Theorem**

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

**Example**

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.

**Example**

Lets

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