Limit of a Function

 How would one prove that the limit of this function, 
$f(x)=x-2$ where $x\neq-2$,
is $L=-4$ at the excluded point?







Lets define $f:(-2,0) \rightarrow R$ by $f(x) = \dfrac {x^2 - 4} {x+2}$. 
    
By definition of a limit, if $x_0$ is an accumulation point of the domain, then $f$ has a limit $L$ at $x_0$ iff for each $\epsilon >0$, there is a $\delta >0$ such that if $0<|x-x_0|<\delta$ then $|f(x) - L|<\epsilon$.


Choose $\epsilon = \delta$
     If $0<|x+2|<\delta$ 
          Then $|f(x) - L| = |\dfrac {x^2 - 4} {x+2} +4| = |(x-2)+4| = |x+2|<\epsilon = \delta$ 

Therefore, $f(x)$ has a limit at $x=-2$, and that limit is $L=-4$.