# How can we ignore the limit point while computing the limit of a function?

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..

PS: I have tried to explain my question here in a better way, as I feel that I didn't make my question clear..

• "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. – Jair Taylor May 7 at 18:42
• 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. – Jair Taylor May 7 at 18:59

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 $$g(x)=\frac{x^2-4}{x-2}$$
• 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.. – thornsword May 7 at 17:48