Assume f is defined for all x near a except possibly at a. Then the lim(x->a+)f(x)=L if snd only if lim(x->a)f(x)=L and lim(x->a-)=L
First, a bit of notation. $f(a+)$ stands for the right-hand limit of $f$ at $a$, likewise $f(a-)$ is the left-hand limit of $f$ at $a$.
In this problem it is given that $f(a-)=f(a+)=L$
Next, if $c>0$ then if $|y-x|<c$ then $x-c<y<x+c$ OR $y\in (x-c,x+c)$. That is $y<x\text{ or }y=x,\text{ or }y<x$.
So if $x\ne y$ then $x$ is to the right of $y$ OR $x$ is to the left of $y$
By definition, if $c>0$ and $f(a+)=L$ means then $\exists d>0$ so that if $a<x<a+d$ then $|f(x)-L|<c$.
Likewise, if $c>0$ and $f(a-)=L$ means that then $\exists d>0$ so that if $a-d<x<a$ then $|f(x)-L|<c$.
What is the definition of $L$ is the $\displaystyle\lim_{x\to a}f(x)=L$ is
- $x\ne a$
- if $c>0$ then $\exists d>0$ so that
- whenever $0<|x-a|<d$ implies $|f(x)-L|<c$
Can you see the each definition implies the other?
no i need to prove in 2 steps so;
lim(x->a+)f(x)=L if lim(x->a)f(x)=L and lim(x->a-)=L
lim(x->a)f(x)=L and lim(x->a-)=L if lim(x->a+)f(x)=L
how would i do this ? sorry not quite understanding
Well you are getting a complete solution from me.Originally Posted by DiscreteMathHelp;922358[COLOR="#FF0000"
I will tell you that what needs doing. You are responsible for doing something.
1) assume that $f(a+)~\&~f(a-)$ both exist. Use those facts and show that the definition of limit at $a$ is satisfied.
2) Assume that the limit at $a$ exists then argue that implies that both $f(a-)~\&~f(a+) must exist.
I gladly read your work and offer help then and only then.