The formal definition of a limit says:
"The function $f$ approaches the limit $L$ near $a$ if $\forall \epsilon>0\ \exists\ \delta>0$ such that $0<|x-a|<\delta \implies |f(x)-L|<\epsilon$"
Taking an example...find the limit: $$\lim_{x \to 2}{\frac{x^2-4}{x-2}}$$
Now, I'm not going to make any guess here, so I cannot invoke the definition directly....nor can I use the standard limit theorem for the quotient of 2 functions because $\lim_{x\to 2}{x-2}=0$
My textbook says, $\pmb{IF}$ $x \neq2$, then we have
$$\lim_{x \to 2}{\frac{x^2-4}{x-2}}=\lim_{x\to2}{x+2}=4$$
So my question here is -- How does ignoring the limit point 2 not affect the answer?..As far as I can see..the formal definition of a limit makes no such statement..in the sense...it's not immediately obvious to me that the $\mathit{definition}$ says
"You can ignore the function's behaviour at the limit point while $\mathit{computing}$ the value of the limit"
Now, I know that the line $0<|x-a|<\delta \implies x\neq a$, but this is only when $\mathit{verifying}$ whether $L$ is the limit of $f(x)$ at $x=a$ or not..(According to the definition..)(I don't see how this is applicable while $\mathit{computing}$ limits)
I'm specifically looking for an answer...based on the definition which allows us to ignore the function's behaviour at the limit point while $\mathit{computing}$ the limit...when the definition makes no such statement..
Thanks for any answers!
PS: I have tried to explain my question here in a better way, as I feel that I didn't make my question clear..
Let $f$ and $g$ be two functions defined on an open interval containing $a$ where $g=f$ except at $x=a$ (we allow the possibility that $g$ is not defined at $a$). Assume that $\lim_{x\to a}f(x)=L$. We claim that $\lim_{x\to a}g(x)=L$. To see this given $\varepsilon>0$ and let $\delta>0$ be such that if $0<|x-a|<\delta$, then $|f(x)-L|<\varepsilon$. But then if $0<|x-a|<\delta$, $f(x)=g(x)$ whence $|f(x)-L|=|g(x)-L|<\varepsilon$ as desired.
In your example $f(x)=x+2$ and $g(x)=\frac{x^2-4}{x-2}$