The Limit of a Function

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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.

the limit of a function

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:$\displaystyle \underset{{x\to {{0}^{+}}}}{\mathop{{\lim }}}\,f(x)=1$

the limit

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

Next article

It can be rather tedious to apply the  and  limit test to individual functions. By remembering some basic theorems about limits we can avoid the some of this repetitive work.
We shouldn’t forget that if a limit exists it is always unique.

Next article

It can be rather tedious to apply the  and  limit test to individual functions. By remembering some basic theorems about limits we can avoid the some of this repetitive work.
We shouldn’t forget that if a limit exists it is always unique.

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