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### Course: AP®︎/College Calculus AB>Unit 1

Lesson 12: Confirming continuity over an interval

# Functions continuous on all real numbers

Discover how to determine if a function is continuous on all real numbers by examining two examples: eˣ and √x. Generally, common functions exhibit continuity within their domain. Explore the concept of continuity, including asymptotic and jump discontinuities, and learn how to identify continuous functions in various scenarios.

## Want to join the conversation?

• What is the variable e? Is it a constant?
• 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.
• 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.
• If we were to add a third dimension then the numbers in that third dimension are not considered "real," in a strict cartesian sense.
• 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 }
• 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
• 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)?
• 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.
• Suppose if I wanted to do a rigorous proof for this, how can I do so?
Using deductive proofs? Principle of Mathematical Induction?
• 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.

• 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.
• 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.
• why is f(x) continuous for all real numbers? F(x) can never be negative even though x is negative.
• 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.
• " 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?
• 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.