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 x2 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.
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: $\displaystyle \underset{{x\to {{0}^{-}}}}{\mathop{{\lim }}}\,f(x)=-1$
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 $\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,({{x}^{2}}+1)=2$.
