# How to interpret f(x) tends to g(x)

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## Summary:

How to interpret the phrase "The function f(x) tends to the function g(x) when x tends to x0"
Hi, I have a question, sometimes one sees in exercises or textbooks some phrase like
The function ##f(x)## tends to the function ##g(x)## when x tends to ##x_0##
My question is, is this unambiguously defined in a mathematical way? I mean, when one reads such a thing this could mean that
$$\lim_{x\to x_0} f(x)=\lim_{x\to x_0} g(x)$$
Another way (I think) this can be interpreted is that
$$\lim_{x\to x_0} \frac{f(x)}{g(x)}=1$$
But also, in addition to ##\lim_{x\to x_0} f(x)=\lim_{x\to x_0} g(x)## one can think also to include a more restrictive condition
$$\lim_{x\to x_0} f'(x)=\lim_{x\to x_0} g'(x)$$
And so on..., even you could try to define that both function must have the same Taylor series arround ##x_0## etc...

So is there a unambiguos definition of that phrase or simply something that physicist say that depend completely on the point of view of the person?

S.G. Janssens

## Answers and Replies

fresh_42
Mentor
That doesn't make much sense at all. If ##x \to x_0## then ##f(x)## tends to a single value ##f(x_0)## which is not a function.

Maybe it is meant as ##x\to \pm \infty## in which case it means that ##\lim_{x \to \pm \infty} |f(x)-g(x)|=0##.

S.G. Janssens
Science Advisor
Education Advisor
That doesn't make much sense at all. If ##x \to x_0## then ##f(x)## tends to a single value ##f(x_0)##
Provided ##f## is continuous at ##x_0##.

fresh_42
Mentor
Provided ##f## is continuous at ##x_0##.
Yes, but even if ##f## is not continuous, then it still converges to a point in ##\mathbb{R}\cup \{\pm \infty\}##, which is no function either.

Stephen Tashi
Science Advisor
My question is, is this unambiguously defined in a mathematical way?
In a manner of speaking, it is defined ambiguously in a mathematical way!

I don't know what textbook you are quoting. One given function ##g(x)## cannot "approach" a different function. There can be situations like ##lim_{x\rightarrow \infty} g(x) = lim_{x \rightarrow \infty} f(x)## but this does not say that ##g(x)## and ##f(x)## "approach" each other at all values.

The context in which functions "approach" other functions is when we are talking about a sequence of functions or parameterized family of functions. For example, you can think of the expression ## f(x,a) = e^{ax}## as defining a parameterized family of exponential functions.

We can speak of a limit of a sequence of functions being a specific function. However "the limit" of a sequence of functions is an ambiguous phrase because there are several distinct definitions for what it means for a sequence of functions to converge to another function. Examples of distinct limit concepts for sequences of functions are "pointwise convergence", "convergence in L2", "uniform convergence". These are often discussed in the context of random variables in probability theory where a limit of a sequence of random variables often amounts to finding a limit of a sequence of their probability distributions.

For a parameterized family of functions ##f(x,a)##, a text may treat ##f## as a function of two variables and speak of ##lim_{a \rightarrow a_0} f(x,a)## as defining a function. As a statement about functions, I interpret this to mean "pointwise convergence".

S.G. Janssens
Science Advisor
Education Advisor
Yes, but even if ##f## is not continuous, then it still converges to a point in ##\mathbb{R}\cup \{\pm \infty\}##, which is no function either.
No, this is not true. If ##f## is not continuous at ##x_0##, then ##f(x)## need not converge at all as ##x \to x_0##.

fresh_42
Mentor
No, this is not true. If ##f## is not continuous at ##x_0##, then ##f(x)## need not converge at all as ##x \to x_0##.
... in which case "tends to" isn't even defined. The question implicitly requires the existence of a limit, for otherwise it wouldn't make any sense either.

S.G. Janssens
Science Advisor
Education Advisor
... in which case "tends to" isn't even defined. The question implicitly requires the existence of a limit, for otherwise it wouldn't make any sense either.
This is opaque and besides the issue.

You made a mistake in post #2.
I briefly pointed it out.
There is nothing more to it.

member 587159
fresh_42
Mentor
You made a mistake in post #2.
I briefly pointed it out.
There is nothing more to it.
Yes, I have assumed a property which wasn't given. But neither of your objections makes ##f(x)## tend to ##g(x)##. Hence the core statement, that if at all, then it is a number and no function, remains valid.

pasmith
Homework Helper
I think "is asymptotic to" is more useful than "tends to" here.

This requirement comes down to $$\lim_{x \to x_0} \frac{f(x) - g(x)}{g(x)} = 0,$$ which if $f$ and $g$ are continuous is equivalent to $$\lim_{x \to x_0} \frac{f(x)}{g(x)} = 1.$$

PeroK
Svein
Science Advisor
You have to select a norm for the function space. The obvious one (to me) is to use a Hilbert space inner product of the type $<f, g> = \int \lvert f\rvert \cdot \lvert g \rvert$ and define the distance d(f, g) as $\sqrt{<f-g, f-g>}$.

I think physicist use this language to denote what mathematicians call an "Equivalence class", that is ##f(x) \sim g(x)## as ##x \rightarrow x_0## but it's been awhile since I've thought about those structures, but that's how I view it.

pasmith
Homework Helper
It should also be noted that $\lim_{x \to x_0}(f(x) - g(x)) = 0$ can hold in circumstances where neither $\lim_{x \to x_0} f(x)$ or $\lim_{x \to x_0} g(x)$ exist separately.

For example, it would make sense to say that $x(1 - e^{-x})$ "tends to" $x$ as $x \to +\infty$.

FactChecker
Indeed, f(x)-g(x) and f(x)/g(x) can have different behaviours when at x0, either both f(x) and g(x) approach zero or both approach infinity.