##### The limit of a function

Let f be a function and let c be a real number. We do not require that f be defined ar c but we do require that f be defined at least on a set of the form (c-p,c) U (c,c+p) with p>0).

To say that $\displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\,f(x)=l$ is to say that $\displaystyle \left| {f(x)-l} \right|$ can be made arbitrarily small simply requiring that $\displaystyle \left| {x-c} \right|$ be sufficiently small but different from zero or you pick $\displaystyle \varepsilon>0$, $\displaystyle \left| {f(x)-l} \right|$ can be made less than $\displaystyle \varepsilon>0$ simply by requiring that $\displaystyle \left| {x-c} \right|$ satisfy an inequality of the form $\displaystyle 0<\left| {x-c} \right|<\delta$ for $\displaystyle \delta$ sufficiently small.

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 $\displaystyle \left| {f(x)-l} \right|<\varepsilon$

Note:

The choice of $\displaystyle \delta$ depends upon the previous choice of$\displaystyle \varepsilon>0$, We do not require that there exists a number $\displaystyle \delta$ which works for all $\displaystyle \varepsilon$ but, rather, that for each $\displaystyle \varepsilon$ there exist a $\displaystyle \delta$ which works for it.

For a $\displaystyle \delta$ to be suitable, all the points within $\displaystyle \delta$ of c must be taken by the function to within $\displaystyle \varepsilon$ of l.

In the figure below we choose $\displaystyle \delta$ too large for the given $\displaystyle \varepsilon$. The points marked x1  and x are not taken by the function to within $\displaystyle \varepsilon$ of  l.

Example: Show that $\displaystyle \underset{{x\to 3}}{\mathop{{\lim }}}\,(2x-1)=5$

Solution: Finding a $\displaystyle \delta$.

Let $\displaystyle \varepsilon>0$. We seek a number $\displaystyle \delta >0$ such that, if $\displaystyle 0<\left| {x-3} \right|<\delta$  then $\displaystyle \left| {(2x-1)-5} \right|<\varepsilon$.

What we have to do first is establish a connection between $\displaystyle \left| {(2x-1)-5} \right|$ and $\displaystyle \left| {x-3} \right|$

The connection is simple:

$\displaystyle \left| {(2x-1)-5} \right|=\left| {2x-6} \right|$ so that $\displaystyle \left| {(2x-1)-5} \right|=2\left| {x-3} \right|$.

To make $\displaystyle \left| {(2x-1)-5} \right|$ less than $\displaystyle \varepsilon$, we need only make $\displaystyle \left| {x-3} \right|$ twice as small. This suggests that we choose  $\displaystyle \delta =\frac{1}{2}\varepsilon$

Proof: Showing that the $\displaystyle \delta$ works.
If $\displaystyle 0<\left| {x-3} \right|<\frac{1}{2}\varepsilon$, then $\displaystyle 2\left| {x-3} \right|<\varepsilon$ and, by $\displaystyle \left| {(2x-1)-5} \right|=2\left| {x-3} \right|$, we have $\displaystyle \left| {(2x-1)-5} \right|<\varepsilon$

### Be careful!

The limit statements $\displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\,f(x)=l$  and $\displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\,\left| {f(x)} \right|=\left| l \right|$ are not equivalent. While the first statement implies the second, the second statement does not imply the first.

We come now to the $\displaystyle \varepsilon$, $\displaystyle \delta$ definitions of one-sided limits.These are just the usual $\displaystyle \varepsilon$,$\displaystyle \delta$statement,except  that, for a left-hand limit, the $\displaystyle \delta$ has to work only for the left of c and for a right-hand limit, the $\displaystyle \delta$ has to work only to the right of c.

### Definition: “Left-Hand Limit”

Let f be a function defined at least on an interval of the form (c-p,c). $\displaystyle \underset{{x\to {{c}^{-}}}}{\mathop{{\lim }}}\,f(x)=l$ if for each $\displaystyle \varepsilon>0$ there exists $\displaystyle \delta >0$ that if $\displaystyle c-\delta <x<c$, then $\displaystyle \left| {f(x)-l} \right|<\varepsilon$

### Definition: “Right-Hand Limit”

Let f be a function defined at least on an interval of the form (c,c+p).$\displaystyle \underset{{x\to {{c}^{+}}}}{\mathop{{\lim }}}\,f(x)=l$ if for each $\displaystyle \varepsilon>0$ there exists $\displaystyle \delta >0$ that if $\displaystyle c<x<c+\delta$, then $\displaystyle \left| {f(x)-l} \right|<\varepsilon$.

One-sided limits give us a simple way of determining whether or not a two sided limit exists:

$\displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\,f(x)=l$ if $\displaystyle \underset{{x\to {{c}^{-}}}}{\mathop{{\lim }}}\,f(x)=l$  and $\displaystyle \underset{{x\to {{c}^{+}}}}{\mathop{{\lim }}}\,f(x)=l$

Example 1: If $\displaystyle f(x)=~\left\{ \begin{array}{l}-1~~~~~~~~x<0\\~~~1~~~~~~~~x>0\end{array} \right.~~$, then $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,f(x)$ does not exist.

SolutionOne-sided limits give us a simple way of determining whether or not a two sided limit exists.

Let’s see if these two sided limits are the same:

The left-hand limit is: $\displaystyle \underset{{x\to {{0}^{-}}}}{\mathop{{\lim }}}\,f(x)=-1$

The right-hand limit is:$\displaystyle \underset{{x\to {{0}^{+}}}}{\mathop{{\lim }}}\,f(x)=1$

Since these one-sided limits are different, $\displaystyle \underset{{x\to {{0}^{+}}}}{\mathop{{\lim }}}\,f(x)$ does not exist.

Example: Decide whether or not the $\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,({{x}^{2}}+1)$exists.

Solution: One-sided limits give us a simple way of determining whether or not a two sided limit exists.

Let’s see if these two sided limits are the same.

The left-hand limit is: $\displaystyle \underset{{x\to {{1}^{-}}}}{\mathop{{\lim }}}\,({{x}^{2}}+1)=2$

The right-hand limit is:$\displaystyle \underset{{x\to {{1}^{+}}}}{\mathop{{\lim }}}\,({{x}^{2}}+1)=2$
Since these one-sided limits are the same, $\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,f(x)$ does exist and $\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,({{x}^{2}}+1)=2$.