2

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

Simply Beautiful Art
  • 74,685
zeraoulia rafik
  • 7,055
  • 1
    If you don't have continuity at $a$, you have absolutely no guarantee that $f(x) \xrightarrow[x\to a]{} f(a)$ (not even that a limit exists). For instance, let $f$ be defined on $\mathbb{R}$ by $f(0)=0$ and $f(x) = \frac{1}{x}\sin \frac{1}{x}$ for $x\neq 0$. – Clement C. Dec 29 '16 at 20:31
  • 3
    For $f(\infty)$ to be defined, the domain must include $\infty$, wich can be done in various ways, but the usual domains $\mathbb R$ and $\mathbb C$ don't and thus there is no $f(\infty)$ in these usual contexts. – AlexR Dec 29 '16 at 20:32
  • See my answer to this question here for the difference between limit and image. –  Dec 29 '16 at 20:58
  • @AlexR, May you meant the titled formula is true for $\bar{R}$ – zeraoulia rafik Dec 29 '16 at 21:06
  • 1
    @user51189 No, the same caveat will apply. $f(\lim)\neq \lim f$ in general. And $\lim f$ may not even exist. – Clement C. Dec 29 '16 at 21:13
  • IMHO it's just convention. From the literature I've read, I think if $\lim_{x\to+\infty} f(x)$ exists in the first place, then $f(+\infty)$ is just a shorthand for this limit, otherwise $f(+\infty)$ simply makes no sense. Same with $f(-\infty),f(\infty)$ etc. The general rule of thumb is, before playing with notations, one must first clarify what they really mean. – Vim Jan 05 '17 at 13:47
  • see the bellow answer by :Mikhail Katz – zeraoulia rafik Jan 05 '17 at 13:49

2 Answers2

0

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.

  • As an aside, another useful thing is the "projective real numbers" which is basically the same thing as the extended reals, except that we make $\infty = -\infty$ (e.g. think of the number line being rolled up into a circle, so that we complete the circle by adding just one point at infinity). This lets us do some additional things, like defining $\frac{1}{0} = \infty$ and $\tan(\frac{\pi}{2}) = \infty$ that would be discontinuous if we were using the extended reals. –  Jan 06 '17 at 09:19
-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.

Mikhail Katz
  • 42,112
  • 3
  • 66
  • 131