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My textbook (Calculus Early Transcendentals, 8th edition, by James Stewart ) asks to verify a function is continuous at a point using the definition of continuity and the limit laws. However, why would the text explicitly state to use the limit laws to verify continuity? I'm assuming this implies that direct substitution shouldn't be used and looking at solutions online confirms my suspicions as they all use limit laws as well. However, using direct substitution is much faster than using the limit laws and both methods achieve the same result.

For example, show that $f(x) = (x+2x^3)^4$ is continuous at $x=-1$

By definition of continuity, we're trying to show that $\lim_{x\to-1}(x + 2x^3)^4 =f(-1)$

Show that $f(-1)$ exists

$$f(-1)=(1+2(-1)^3)^4 = 81$$

Now for the limit

Using the limit laws:

$$\lim_{x\to-1}(x + 2x^3)^4$$

$$=[\lim_{x\to-1}x + 2x^3]^4$$

$$=[\lim_{x\to-1}x + \lim_{x\to-1}2x^3]^4$$

$$=[\lim_{x\to-1}x + 2\lim_{x\to-1}x^3]^4$$

$$=[-1 + 2(-1)^3]^4 = 81$$

Using direct substitution: $$\lim_{x\to-1}(x + 2x^3)^4 = (1+2(-1)^3)^4=81$$

In either method we reach the same result that $\lim_{x\to-1}(x + 2x^3)^4 = 81$ which verifies that $f(x)$ is continuous at $x=-1$ since $\lim_{x\to-1}(x + 2x^3)^4 =f(-1)$

Perhaps I'm missing some important connection between continuity and limits as to why limit laws are used instead of direction substitution to verify continuity?

Slecker
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    Consider $\lim\limits_{x\to 0}(\lceil x^2\rceil) = 1 \neq \lceil 0^2\rceil$. "Direct substitution" assumes continuity and cannot verify continuity without being circular logic. – JMoravitz Jul 26 '20 at 15:10
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    In this case your function is continuous and direct substitution will work. However, in case of rational functions, floor, ceiling functions etc. limits laws are necessary. – SarGe Jul 26 '20 at 15:11
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    Yeah, what exactly is meant by "direct substitution"? Aren't you just plugging in the number ${-1}$ once again and showing ${f(-1)=81}$? In any case - limit laws allow you conclude things about limits, which is what we need to do to verify continuity. Directly substituting in the number for the limit will assume continuity. This example is simple, and so it doesn't really matter. But for other examples - not so much. Just as @JMoravitz has also said. Hope that helps :) – Riemann'sPointyNose Jul 26 '20 at 15:14
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    @Riemann'sPointyNose Yes, that's about how my textbook defines direct substitution: "If $f$ is a polynomial or a rational function and $a$ is in the domain of $f$, then $\lim_{x\to a}f(x) = f(a)$." Ah I see, so if we know that continuity occurs at a point then "direct substitution" is allowed. Otherwise, we use the limit laws to verify continuity. I think I have a better grasp now. Thank you all! – Slecker Jul 26 '20 at 15:22
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    @Slecker correct. You're book is basically saying that if ${f}$ is a polynomial or rational function and ${a}$ is in it's domain, then it's continuous at that point. But in trying to prove continuity you cannot use this fact (since you'd be saying "it's continuous, because it's continuous"). No problem! – Riemann'sPointyNose Jul 26 '20 at 15:27
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    Also - just did an edit to my answer to show another way to show continuity of ${(x+2x^3)^4}$! :) @Slecker – Riemann'sPointyNose Jul 26 '20 at 15:51

1 Answers1

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Maybe it'll help to break down very strictly what we are doing with the limit laws. Firstly,

$${(x + 2x^3)^4 = (x+2x^3)(x+2x^3)(x+2x^3)(x+2x^3)}$$

Now, if we have that the limit

$${\lim_{x\to -1}(x+2x^3)}$$

exists, we can say that

$${\lim_{x\to -1}(x+2x^3)^4 = (\lim_{x\to -1}(x + 2x^3))(\lim_{x\to -1}(x + 2x^3))(\lim_{x\to -1}(x + 2x^3))(\lim_{x\to -1}(x + 2x^3))=(\lim_{x\to -1}(x + 2x^3))^4}$$

this is because the limit of the product is the product of the limits (provided the individual limits exist). This is indeed a limit law.

We know that both ${x}$ and ${2x^3}$ will be continuous functions, and so will their sum. Hence we can say that

$${\lim_{x\to -1}x+2x^3=-1+2(-1)^3=-3}$$

Notice we can only directly substitute ${x=-1}$ because we know ${x+2x^3}$ is continuous. This step becomes invalid if we did not know the continuity of ${x+2x^3}$. And now since the individual limit of ${x+2x^3}$ exists, we have that

$${\lim_{x\to -1}(x+2x^3)^4 = (-3)^4 = 81}$$

And ${f(x)=(x+2x^3)^4}$ also has ${f(-1)=81}$, we can conclude ${f(x)}$ is continuous at ${x=-1}$.

If you just directly plug in the number ${-1}$ into

$${(x+2x^3)^4}$$

and claiming this is

$${\lim_{x\to -1}(x+2x^3)^4}$$

you are assuming that ${(x+2x^3)^4}$ is continuous. So you cannot use this to prove continuity, as this is circular reasoning, as also stated by @JMoravitz

Edit: another way to show continuity is to notice that if ${f(x) = x^4}$ and ${g(x) = x + 2x^3}$, notice that your function is just

$${(x+2x^3)^4 = f(g(x))}$$

Both ${f,g}$ are continuous, and the composition of two continuous functions is continuous - so ${(x+2x^3)^4}$ is continuous

Riemann'sPointyNose
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