Is this :$\displaystyle\lim_{x\to \infty} f(x) =f(\infty) $ true?

Question

I would like to know the difference between limit and image of function. For example, if I define the function by $f(x)= x^2$ , the image of the function by $x=2$ is $f(2)=2^2=4$ and if i would like to calculate $\displaystyle\lim f(x) ,x\to 2$ we w'll obtain $4 =f(2)$ in this case Image has the same mathematical meaning with limit . My humble question here is :

Question: What is the mathematical difference between limit and Image of function and have they the same meaning for $x=\infty$ ?.

Note: I know only they have the same meaning if $f$ is a continuous function .

Thank you for any help

Answer 1

This is almost correct if interpreted in the context of a hyperreal extension $\mathbb R\hookrightarrow{}^\ast\!\mathbb R$. Namely, if $H$ denotes an infinite hyperreal then the limit $\lim_{x\to\infty}f(x)$ will exist if and only if for every $H$ the value of $f(H)$ is infinitely close to a suitable real $L$ (independent of the choice of $H$). Then $$ \lim_{x\to\infty}f(x)\approx f(H) $$ where $\approx$ stands for the relation of infinite proximity.

Answer 2

Just like any other value, $\lim_{x \to \infty} f(x) = f(\infty)$ holds precisely when $f$ is continuous at $\infty$.

The use $\infty$ and $-\infty$ in calculus can be made into actual values one can compute with; adding these two points to the ends of the real line gives something we call the "extended real numbers".

We usually, we only bother to define the value of a function at $\infty$ when doing so makes the function continuous — e.g. we define $\arctan(\infty) = \frac{\pi}{2}$ and $\arctan(-\infty) = -\frac{\pi}{2}$, but we leave $\sin(\infty)$ undefined.

In other words, the limit at and the value at $\infty$ always the same is the result of convention. If we don't stick to that convention, then we have to be prepared for situations where $f(\infty)$ exists, but $\lim_{x \to \infty} f(x)$ has a different value, or doesn't exist at all.