lim x→2g(x)
whereg(x) = ( x − 2 /x^ 2 − 4) if x ≠ 2
g(x)= 1 if x = 2
this confuses me any help would be nice ....
As it is written, your function is $\displaystyle \begin{align*} x - \frac{2}{x^2} - 4 \end{align*}$, but I suspect you meant $\displaystyle \begin{align*} \frac{x - 2}{x^2 - 4} \end{align*}$. Which one is it? If it's the latter, you MUST use brackets where they are needed!
NEVER, NEVER, NEVER Say that. It will earn you a failing mark.
The notation $x\approx a$ means that $x$ is close to $a$.
So $\displaystyle \lim_{x\to a}f(x)=L$ means that if $x\approx a$ but $x\ne a$ then $f(x)\approx L$.
Now $\dfrac{x-2}{x^2-4}=\dfrac{1}{x+2}$ unless $x=2$.
So $\displaystyle \lim_{x\to 2}\dfrac{x-2}{x^2-4}=\displaystyle \lim_{x\to 2}\dfrac{1}{x+2}=\dfrac{1}{4}$ because by definition of limit $x\ne 2$ so $\dfrac{x-2}{x^2-4}=\dfrac{1}{x+2}$
$\dfrac{x-2}{x^2-4}$ is not defined at $x=2$ because you are dividing by zero (which is not a valid operation). However, they defined $g(x) = \begin{cases}\dfrac{x-2}{x^2-4} & x\neq 2 \\ 1 & x=2\end{cases}$ to mean that if $x\neq 2$, then $g(x) = \dfrac{x-2}{x^2-4}$ and $g(2) = 1$.
A limit of a function has nothing to do with the value of that function at that point. This is what is known as a removable discontinuity.
I don't know what you mean by "imply a different part of the graph". The graph of y= g(x) is the graph of 1/(x+ 2) except that the point (1, 1/4) is removed and the point (1, 1) is added. But you take x closer and closer to 1, but not equal to 1, (x, y) on the graph gets closer and closer to (1, 1/4).