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## Calculus, all content (2017 edition)

# Functions continuous on all real numbers

AP.CALC:

LIM‑2 (EU)

, LIM‑2.B (LO)

, LIM‑2.B.1 (EK)

, LIM‑2.B.2 (EK)

Sal is asked which of the following two functions is continuous on all real numbers: eˣ and/or √x. In general, the common functions are continuous on all the numbers

*in their domain*.## Want to join the conversation?

- What is the variable e? Is it a constant?(20 votes)
- e stands for Euler's constant, which is about 2.718. It's a bit like pi: they're both specific symbols to refer to an irrational constant.

e is closely tied to the natural logarithm, which is written as ln(x). In general, log(x) involves a log of base ten, so log(x)=y means that 10^y=x. The natural logarithm is base e, so ln(x)=y means that e^y=x.

e and the natural logarithm are often used when dealing with questions of population growth. They are also very useful in calculus due to their derivative properties. You'll probably cover that later on.

Hope this was helpful.(68 votes)

- Well, isn't g(x) continuous for all real numbers? We just need an additional axis in the third dimension in order to plot the complex numbers. Or we can plot the function on the complex plane. The function will still be continuous for all real numbers x but in the case of the negative real numbers there will still be continuity but the values of the function evaluated at those points will be complex numbers.(7 votes)
- If we were to add a third dimension then the numbers in that third dimension are not considered "real," in a strict cartesian sense.(21 votes)

- According to this, a function is continuous if and only if f(x) as x approaches a = f(a). But what if we have a piecewise function, like,

g(x) = {3x, x does not equal 2}

{-10, x = 2 }(6 votes)- Then it is clearly not continuous because of the removable discontinuity at x=2. We can prove that by using the limit definition of continuity that Sal showed in the video.

f is continuous at a, if and only if lim_(x->a) f(x) = f(a)

Now, for your piecewise function, g(x) = 3x for when x≠2 and g(x) = -10 for when x=2.

Given that g(2) = -10

lim_(x->2) g(x) = lim_(x->2) 3x = 3 * 2 = 6 ≠ g(2) = -10

Since the lim_(x->2) g(x) ≠ g(2) it is not continuous at x=2(12 votes)

- But f(x) = e^x is an exponential function, therefore it has an asymptote. Seeing as the limit as x approaches the asymptote would technically not exist, does that not mean the function isn't continuous? Or does that not count as the asymptote is technically not part of the domain of f(x)?(6 votes)
- It does have and asympote, however as you can see it gets closer and closer to that asymptot as x increases. It is true that it will never reach but as we progress it will get closer and crosser to a number. However, because there are never any gaps as it get closer, no holes or breaks, we can assume that if we were to reach infinity, say it was a real number, then we would touch that asymptote. Therefore it is contiguous. Now if there was a single whole anywhere leading up to that asymptote then it would not be contiguous.(2 votes)

- Suppose if I wanted to do a rigorous proof for this, how can I do so?

Using deductive proofs? Principle of Mathematical Induction?(2 votes)- To talk about continuous functions rigorously, you need rather a lot of background knowledge in real analysis, which I can't condense into an answer here. You can try the book Understanding Analysis by Stephen Abbot, which I've found to be a well-written text, and check the section on continuous functions, where it's defined formally. Anything you'll first need to understand rigorously, like limits of sequences or properties of
**R**, is earlier in the same book.

The book can be downloaded for free, and found with a Google search.(6 votes)

- why is f(x) continuous for all real numbers? F(x) can never be negative even though x is negative.(1 vote)
- Continuity only requires that the function be well-defined at any given point and that the limit at each point is equal to the value of the function at that point. The codomain of the function does not matter.(3 votes)

- Why is g of x not defined for all real numbers? And how do you tell? I can't seem to get the hang of this.(1 vote)
- g(x) = sqrt(x) is not defined for all real numbers because if you take the square root of a negative number you get what mathematicians call an imaginary number. This is because a negative times a negative is always a positive and a positive times a positive is always a positive, meaning that you cannot have a real number times itself equal a negative. Because of this, g(x) is not defined for all real numbers.(2 votes)

- " In essence, these are functions whose graphs can be drawn with a single brush stroke." But one cannot draw f(x)=1/x with one stroke, despite it being a continuous function, right?(1 vote)
- f(x) = 1/x is not defined at x = 0, so it is not continuous for all reals. Moreover, you can't find a value for f(0) that would make the function continuous, so the discontinuity is not removable.

However, to see that the "single brush stroke" analogy is a bit unfortunate anyway, take a look at the https://en.wikipedia.org/wiki/Weierstrass_function.(2 votes)

- What does "In general, the common functions are continuous on all the numbers in their domain", mean? Also, what is e^x and where can I learn more about this?(1 vote)
- A function is continuous if it is defied for all values, and equal to the limit at that point for all values (in other words, there are no undefined points, holes, or jumps in the graph.) The common functions are functions such as polynomials, sinx, cosx, e^x, etc. e^x is the exponential function, https://www.youtube.com/watch?v=6WMZ7J0wwMI e is an important mathematical constant like pi, that shows up a lot when talking about exponential growth.(2 votes)

- Why is g(x)=√x not a continuous function? He says that it's not defined for negative values of x, but √x is always a positive value. You can't take negative value for √x. Please help me out.(1 vote)
- You've just about got it. Because you can't take the square root of a negative number, sqrt(x) doesn't exist when x<0. Since the function does not exist for that region, it cannot be continuous. In this video, we're looking at whether functions are continuous across all real numbers, which is why sqrt(x) is described simply as "not continuous;" the region we're talking about is not specific to the function.(2 votes)

## Video transcript

- [Voiceover] Which of
the following functions are continuous for all real numbers? So let's just remind ourselves what it means to be continuous. What a continuous function looks like. So, a continuous function, let's see, that's my y-axis, that is my x-axis. A function is going to be
continuous over some interval. If it just has, doesn't have any jumps or discontinuities over that, or gaps over that interval, so if it's connected and it for sure has to be
defined over that interval without any gaps, so for example, a continuous function could
look something like this. This function, let me make that line
a little bit thicker, so this function right
over here is continuous. It is connected over this interval, the interval that we can see. Now, examples of discontinuous functions over an interval, or
non-continuous functions, well, they would have gaps of some kind. They could have some type of
an asymptotic discontinuity so something like that, that makes it discontinuous. They could have a jump of discontinuity, something like that. They could just have a gap where they're not defined, so they could have a gap
where they're not defined, or maybe they actually are defined there, but it's removable discontinuity, so all of these are examples
of discontinuous functions. Now, if you want the more
mathy understanding of that and we've looked at this before, we say that a function f is continuous, continuous at some value, x equals a, if and only if, draw my little two-way arrows here, say if and only if the limit of f of x as x approaches a is equal to the value
of the function at a, so once again, in order
to be continuous there, you at least have to be defined there. Now, when you look at these,
the one thing that jumps out at me, in order to be
continuous for all real numbers, you have to be defined
for all real numbers and g of x is not defined
for all real numbers. It's not defined for negative values of x, and so, we would rule this one out, so let's think about f
of x equals e to the x. It is defined for all real numbers, and as we'll see, most of the common functions
that you've learned in math, they don't have these
strange jumps or gaps or discontinuities. Some of them do, functions like 1 over x and things like that, but things like e to the x,
it doesn't have any of those. We could graph e to the x. E to the x looks something like, e to the x looks something like this, it's defined for all real numbers, there's no jumps or gaps of any kind and so, this f of x is continuous for all real numbers and f only. Now, I didn't do a very rigorous proof. You could if you like, but for the sake of this exercise, it's really more of getting
this intuitive sense of like, look, e to the x is
defined for all real numbers and so, and there's no jumps or gaps here so it's reasonable to
say that it's continuous but you could do a more
rigorous proof if you like as well.