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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..

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    $\begingroup$ "I don't see how this is applicable while computing limits..." It seems like you're saying that you only use the definition to prove that a limit exists, and this doesn't matter when you're only trying to find the limit. This is not right. The definition is the definition: if the definition were different, a limit would mean a different thing, and you would have to use a different method in order to find it. $\endgroup$ – Jair Taylor May 7 at 18:42
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    $\begingroup$ As you've pointed out, the definition of $\lim_{x\rightarrow a} f(x) = L$ does not say anything about what happens when $x = a$. It only matters what happens when $x$ is very close, but not equal to, $a$. So, the actual value of $f(a)$ makes no difference. $\endgroup$ – Jair Taylor May 7 at 18:59
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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 $g(x)=\frac{x^2-4}{x-2}$

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  • $\begingroup$ We are not computing the limit here right?...We know that the answer is 4..which is why you have taken $L$ to be 4 in your answer.. $\endgroup$ – thornsword May 7 at 17:48
  • $\begingroup$ See the general revised answer $\endgroup$ – Sri-Amirthan Theivendran May 7 at 18:02

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