# Introduction to Limits

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## Numerical and Graphical approach to limits

Numerical Approach

Let’s take a function f(x) and see how the values of the functions change when x takes values closer to a specific number.

Example: Let f(x)=3x+1 and calculate f(x) as x takes values closer to 1, but not exactly the value at 1.

We first consider values of x approaching 1 from the left (x<1)

We first consider values of x approaching 1 from the right (x>1)

In both cases as we can see when x approaches 1, f(x) approaches 4. So intuitively we say that $\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,f(x)=4$

Graphical Approach

Even in the graph in both cases as we can see as x approaches 1, f(x) approaches 4. So intuitively we say that $\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,f(x)=4$.
If we find f(1) we see that f(1) = 3 x 1 + 1 = 4 that is the same value as the limit. So $\displaystyle f(1)=\underset{{x\to 1}}{\mathop{{\lim }}}\,3x+1=4$. This is true because our function is defined on the point 1.

##### Be careful!

We are explaining about the values that f(x) takes when x gets a lot closer to 1 and not the value of f(1). Because in some cases we can’t find the value of the function on a given point if the function is not defined on that point. But we can find it´s limit even if the function is not defined on that point but approaches it very closely.

Let’s see an example when our function is not defined on the point in which x approaches the function.

Example: Let $\displaystyle f(x)=\frac{{{{x}^{2}}-1}}{{x+1}}$and calculate f(x) as x takes values closer to -1.

We first consider values of x approaching -1 from the right (x<-1)

We first consider values of x approaching -1 from the left (x>-1)

In both cases as we can see when x approaches -1, f(x) approaches -2. So intuitively we say that $\displaystyle \underset{{x\to -1}}{\mathop{{\lim }}}\,\frac{{{{x}^{2}}-1}}{{x+1}}=-2$

In an algebraic approach we conclude the limit this way:

$\displaystyle \underset{{x\to -1}}{\mathop{{\lim }}}\,\frac{{{{x}^{2}}-1}}{{x+1}}=$

$\displaystyle \underset{{x\to -1}}{\mathop{{\lim }}}\,\frac{{\left( {x-1} \right)\left( {x+1} \right)}}{{x+1}}=$

$\displaystyle \underset{{x\to -1}}{\mathop{{\lim }}}\,(x-1)=-1-1=-2$

Graphical Approach

Even in the graph in both cases as we can see as x approaches -1, f(x) approaches -2. So intuitively we say that $\displaystyle \underset{{x\to -1}}{\mathop{{\lim }}}\,\frac{{{{x}^{2}}-1}}{{x+1}}=-2$.We find f(-1)  because our function is not defined at the point -1 since our denominator is 0. $\displaystyle f(-1)=\frac{{{{{(-1)}}^{2}}-1}}{{-1+1}}=\frac{0}{0}$

### What about when it has a different value from differentsides

Let $\displaystyle f(x)=\left\{ \begin{array}{l}x+1~~~~~x>2~~\\x-1~~~~~x<2\end{array} \right.~$and approach graphically f(x) when x gets closer to 2 from the left and the right.

From the graphic we see that when we approach values of x from the right of the value 2,  the value of the first part of our function gets closer to 3 but never 3. Meanwhile when we approach values of x from the left of the value 2, the value of the second part of our function gets closer to 1. So we get on the conclusion that our function has different limits when approached from different sides.

Be careful!
In this case the ordinary limit does not exist since the one-sides limits are not the same. But we can say separately that the right and left limits of this functions exists and we found them.

Next article

### The limit of a function

Definition: “The Limit of a Function”
Let f be a function defined at least on some set of the form (c-p, c) U (c, c+p). $\displaystyle\underset{{x\to c}}{\mathop{{\lim }}}\, f(x)=l$ if for each $\displaystyle \varepsilon>0$ there exists $\displaystyle \delta >0$ such that if $\displaystyle 0<\left| {x-c} \right|<\delta$ for $\displaystyle\delta$ , then $\displaystyle \left| {f(x)-l} \right|<\varepsilon$

Next article

### The limit of a function

Definition: “The Limit of a Function”
Let f be a function defined at least on some set of the form (c-p, c) U (c, c+p). $\displaystyle\underset{{x\to c}}{\mathop{{\lim }}}\, f(x)=l$ if for each $\displaystyle \varepsilon>0$ there exists $\displaystyle \delta >0$ such that if $\displaystyle 0<\left| {x-c} \right|<\delta$ for $\displaystyle\delta$ , then $\displaystyle \left| {f(x)-l} \right|<\varepsilon$