# Homework Help: A limit of F(x) says what we already know?

1. Aug 31, 2010

### Rhine720

I've been getting very annoyed with learning limits lately. Through direct substitution we can find the limit of something, I know that.

F(x)= x + 2. Intuitively, I know that if I were to substitute 2 to x I would get 4.
F(4)= 2+ 2 = 4. This appears to be correct.

Likewise, the LIMIT of F(x) AS X APPROACHES 2. F(x) = x+ 2. I'm going to use direct substitution. F(2)=2+2=4.

Does F(x)=Limit?

Also, I was under the wrong idea that it meant the limit was that number so you could never reach that number. So I also came to think well then this is telling me I can put in any number that's very close to 2 and I will get a number very close to 4. This is too obvious to be true. F(x)= 2 + 2 , F (1.999)= 1.999 + 2 = 3.999

2. Aug 31, 2010

### cepheid

Staff Emeritus
Sure, you're considering trivial cases right now. If the function is defined at that point, then the limit can be obtained by direct substitution. The point is that the limit can still be defined approaching that value even when the function is NOT defined at that value (in which case you definitely CAN'T evaluate it by direct substitution!). Furthermore, sometimes the limit can be different approaching from one direction than from another (e.g. if you have a jump discontinuity).

This very precise and seemingly tedious wording that lim (x-->a) of f(x) = b means that the function can be made arbitrarily close to b by making x arbitrarily close to a (where "arbitrarily close" means, "as close as you want"), is actually quite important in other situations. For instance, how else would you interpret the concept of a limit "at infinity?"

It is often the case in math that you are given precise expressions of new concepts and ideas BEFORE being exposed to applications that would give you a sense of some of the motivation for introducing those ideas. As you continue to study limits and see some of their applications, it will become clearer why they exist.

3. Aug 31, 2010

### Office_Shredder

Staff Emeritus
This is in fact exactly the idea of a limit. When you have a continuous function asking what the limit is seems stupid because the function will always take the value of the limit. But if you have a function like

$$\frac{sin(x)}{x}$$

then what's the limit as x goes to zero? You can't just plug in 0. That's where limits are useful to describe what the function looks like

4. Sep 1, 2010

### HallsofIvy

Be careful here. Many students get the wrong impression that "limit" is just a fancy way of talking about the value of a function. That is only true for the "trivial cases" cepheid is talking about. Specifically, we say that a function is continuous at a point if and only if the limit exist and is the value of the function. Almost all the function we work with in basic Calculus are continuous- but that is because we like "simple" functions! In fact "almost all" functions (in a very specific sense) are not continuous at any point. But because continuous functions are so easy to work with, almost all of the functions we use are continuous.

5. Sep 1, 2010

### nobahar

Hello!
I was just wondering if someone could elaborate on this.

6. Sep 1, 2010

### jgens

I'm new to this type of stuff, so I'm not sure how well this works, but I have an idea. I think that the problem reduces to showing that the cardinality of the set of continuous functions is strictly less than that of the discontinuous functions. So here are my thoughts ...

A function f:RR is simply a collection of ordered pairs with the property that if (a,b) and (a,c) are are in f, then b=c. Let f be a continuous function, in which case it's clear that each ordered pair of f can be put into a one-to-one correspondence with a unique member of R, and thus |f|=|R|. Now consider P(f) (the powerset of f). Each member of P(f) is a subset of f, and therefore satisfies the criterion for being a function (except perhaps the empty set).* Since f is the only possible member of P(f) which is continuous on R, the set P(f)/{Ø,f} contains only discontinuous functions; moreover, we clearly have that |P(f)/{Ø,f}|=|P(R)|. Therefore, the set of discontinuous functions has cardinality of at least |P(R)|, while the set of continuous functions only has cardinality |R|.**

**http://en.wikipedia.org/wiki/Cardinality_of_the_continuum#Sets_with_cardinality_c

7. Sep 1, 2010

### Office_Shredder

Staff Emeritus
The problem with that argument is that f is the only element of P(f) which is even a function on R, let alone continuous. The proper subsets of f are just continuous functions on proper subsets of R.

However, functions from R to R are clearly at least as many as the functions from R to {0,1}. P(R) is in bijection with this set by if A is a subset of P(R), A corresponds to the function where f(x)=1 if x is in A, 0 otherwise

8. Sep 1, 2010

### HallsofIvy

Consider the function
$$f(x)= \begin{bmatrix} 2x if x\ne 0 \\ 5 if x= 0$$

The value of f(0) is 5 but $\lim_{x\to 0} f(x)= 0$ because if we are "almost" at x= 0 but not exactly (which is what the limit talks about), say $x= \delta$ which is just slightly different from 0, then $f(x)= 2\delta$ which is almost 0. As $x= \delta$ gets closer and closer to 0 (but not at 0) $f(x)= 2\delta$ gets closer and closer to 0.

The most important "law of limits", often not given the attention it deserves is "If f(x)= g(x) for all x except x= a, then $\lim_{x\to a} f(x)= \lim_{x\to a} g(x)$." In fact, the value of f(a) is completely irrelevant to the value of $\lim_{x\to a} f(x)$.

9. Sep 1, 2010

### jgens

Again, I'm new to this stuff, but couldn't this be addressed by the following argument?

Let f(x)=c and consider P(f). For each function g in P(f) which is only defined on some proper subset of R, we can define a corresponding function h such that h(x)=g(x) for all x in the domain of g and h(x)=c+1 for all real x not in the domain of g. If A is the set of these functions h, it should follow that |A|=|P(f)\{Ø,f}| and moreover, every member of A should be discontinuous and defined on R.

10. Sep 4, 2010

### Rhine720

I've been looking in my book lately on discontinuitys and I realized that a number that can't possibly be in the domain of the function is what leads to holes in the range of the function.

So when I look at things like these, assuming I can't factor my way around this so that there actually is an x and y for the limit, I have to choose a number arbitrarily close to it?

And also that a limit that fails to exist is one with en entire jump with no points to fill it in or anything.

I'm playing with my calculator, taking a more numerical approach to understanding this. I know f(x)=1/x as x approaches 0 has no y for a x of zero. But I know it must be infinity for -.999 for x and 1 for x yield pretty much the same results.

Yes?