Exercise 6.1.9

Exercise statement

Explain why Theorem 6.1.19(f) fails when the limit of the denominator is $0$ . (To repair that problem requires L’Hôpital’s rule, see Section 10.5.)

Hints

None.

How to think about the exercise

This is a straightforward exercise, so I don’t have anything to say.

Model solution

The limit of the denominator is $y$ . If $y=0$ , then $x/y$ is not defined, so it doesn’t make sense to say that $(a_n/b_n)_{n=m}^\infty$ converges to $x/y$ .

Could $(a_n/b_n)_{n=m}^\infty$ still converge to something? In other words, does the limit $\lim_{n\to\infty} (a_n/b_n)$ exist? It can sometimes exist. For instance, if we have $a_n := 1/n$ and $b_n := 1/n$ , then $b_n \to 0$ as $n\to \infty$ . But also $a_n/b_n = 1$ for all $n$ , so $a_n/b_n \to 1$ as $n \to \infty$ , so the limit exists. In this case $x/y = 0/0$ which is not defined. If instead we have $a_n := 1$ for all $n$ , then $a_n/b_n = n$ so the sequence diverges. Thus $(a_n/b_n)_{n=m}^\infty$ does not always converge either.

2 thoughts on “Exercise 6.1.9”

Can I argue by contradiction ?
That is I assume for the sake of contradiction that f) holds, then we have $\lim_{x \to \infty}(a_n/b_n)=x/0$ , which from c) implies $b_n=0$ for all $n\geq m$ , contradicting the given.

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Issa Rice says:

April 29, 2023 at 6:19 pm

You can only use part (c) if you already know the sequence is constant. But the more basic point is that once you’ve written $x/0$ you’re in trouble because that’s not a number.

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