Use the $\epsilon$,$\delta$-definition of limits to show that $\lim\limits_{x\to a} f(x)=L$ if and only if $\lim\limits_{x \to a^+} f(x)=\lim\limits_{x\to a^-}f(x)=L$
I started by defining the left- and right-hand limit respectively. and then I assume that the limit is $L$ when $f(x)$ goes to $a$. but I have no idea how to continue. can anyone help? thanks and much appreciated.
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$\begingroup$You can use the delta epsilon method of proving this (which I assume is what you want) by the following.
Right-handed limit (when $x>a$)
There exists $\delta$ such that if $x-a < \delta$, $|f(x)-L|<\epsilon$
Left-handed limit (when $x<a$)
There exists $\delta$ such that if $a - x < \delta$, $|f(x)-L|<\epsilon$
The two-sided limit is defined as : There exists $\delta$ such that if $|x-a| < \delta$, $|f(x)-L|<\epsilon$
We can split up the absolute value around $x-a$ above on the two sided limit in the following way:
If $x-a>0$, or equivalently if $x>a$
$|x-a| < \delta$ Becomes $x-a < \delta$ The same as the right sided limit definition.
If $x-a<0$, or equivalently if $x<a$
$|x-a| < \delta$ Becomes $-(x-a) < \delta$
$a - x < \delta$
The same as the left sided limit definition.
This means that, in order for the two sided limit to exist, both the right handed and left handed limits must exist.
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