As far as I understand, the following is my personal definition for limits/limiting-values:
If the value of a function cannot be determined for any value of the independent variable, then, the value the function seems to be approaching would be its limiting value for that particular value of the independent variable.
For instance, $f(x)=\frac{2x}{x+1}$'s limiting value is $2$ at $\infty$ as we can't determine the value of $f(x)$ at $x=\infty$. But, it seems that $f(x)$ is approaching $2$ when $x$ approaches $\infty$.
Am I correct?
$\endgroup$1 Answer
$\begingroup$With reference to your first claim "If the value of a function cannot be determined for any value of the independent variable $\ldots$", as noticed in the comments by T.Bongers, recall that the limit at a point has nothing to do with the existence of a value for the funcion at that point.
For example we know that $\frac{\sin x}{x}$ is not defined at $x=0$ and $\lim_{x\to 0}\frac{\sin x}{x}=1$ holds even if we define, for example, $\frac{\sin x}{x}=\pi$ at $x=0$.
When at a point $x=x_0\in \mathbb{R}$ the limit value coincides with the value of the funtion at that point we define the function continuous at that point, that is
$$\lim_{x\to x_0} f(x)=f(x_0) \iff \text{f is continuous at x=x_0}$$
With reference to the example, your rough idea "(seems) that $f(x)$ is approaching 2 when $x$ approaches $\infty$" is correct, even if not rigorous of course.
More precisely we could say that the function approaches $2$ closer and closer as $x$ becomes larger and in a rigorous way we say that
$$\forall \epsilon >0 \quad \exists \bar x \quad \forall x>\bar x \quad |f(x)-2|<\epsilon$$
that is
- for any fixed "small" number $\epsilon >0$
- (at least) a value $\bar x$ exists such that for any $x$ larger than $\bar x$ (i.e."eventually")
- $f(x)$ approches the value $2$ with a difference less than $\epsilon$
