Given that $$\lim_{x\to0^+}\frac{1}{x}=\infty$$Is it possible to prove that $$\lim_{x\to0}\frac{1}{x^2}=\infty$$ using only the basic limit laws (such as the "limit of a sum is the sum of limits", and the analogous for subtraction, multiplication, division) and, if necessary, the fact that $\lim_{x\to a}f(g(x))=f(\lim_{x\to a}g(x))$ if $f(x)$ is continuous at $x=\lim_{x\to a}g(x)$?

I'm trying to explain this to a friend, by my reasoning depends on saying that $$\lim_{x\to0^+}\frac{1}{x^2} =\left(\lim_{x\to 0+ }\frac{1}{x}\right)^2$$ which I can't say if its a legal step, since $x^2$ isn't "continuous at infinity".

  • $\begingroup$ Technically, you are right. $f(x)=x^2$ is the function you are composing with $g(x)=\frac{1}{x}$. There is a sense in which $f(x)=x^2$ is continuous at infinity, but it is not a beginning calculus definition. $\endgroup$ – Thomas Andrews Mar 27 '15 at 21:52
  • $\begingroup$ Sorry, who's "he" in your comment? $\endgroup$ – JLagana Mar 27 '15 at 21:53
  • $\begingroup$ Sorry, misread who objected. $\endgroup$ – Thomas Andrews Mar 27 '15 at 21:54
  • $\begingroup$ Oh, ^^. Well, is there a way to prove this without resorting to a "not a beginning calculus definition"? $\endgroup$ – JLagana Mar 27 '15 at 21:56
  • $\begingroup$ Squaring is another way of saying "multiply this by itself". In other words, instead of applying the "square" function to the limit, why not just multiply it by itself? $\endgroup$ – pjs36 Mar 27 '15 at 22:00

If you agree that

$$\lim_{x\to0^+}{1\over x}=\lim_{x\to0}{1\over |x|}$$

then, using an appropriate "multiplication" theorem for limits of the form $\lim_{x\to a}(f(x)g(x))=\lim_{x\to a}(f(x))\lim_{x\to a}(g(x))$, we have

$$\lim_{x\to0}{1\over x^2}=\lim_{x\to0}{1\over|x|^2}=\left(\lim_{x\to0}{1\over|x|} \right)^2=\left(\lim_{x\to0^+}{1\over x} \right)^2=\infty$$

Added later (at OP's request): All you really need here is a theorem that says

$$\lim_{x\to a}f(x)=\lim_{x\to a}g(x)=\infty\implies \lim_{x\to a}(f(x)g(x))=\infty$$

But this follows easily from the definition of $\infty$ as a limit: $\lim_{x\to a}f(x)=\infty$ if for all $M\gt0$ (no matter how large) there exists a $\delta\gt0$ such that $|x-a|\lt\delta\implies f(x)\gt M$. Here's the proof: Let $M\gt0$, and let $M=M_1M_2$ with $M_1,M_2\gt0$. Since $\lim_{x\to a}f(x)=\infty$, there is a $\delta_1\gt0$ such that $|x-a|\lt\delta_1\implies f(x)\gt M_1$. Likewise, since $\lim_{x\to a}g(x)=\infty$, there is a $\delta_2\gt0$ such that $|x-a|\lt\delta_2\implies g(x)\gt M_2$. Now let $\delta=\min(\delta_1,\delta_2)$. Then $|x-a|\lt\delta$ implies $f(x)\gt M_1$ and $g(x)\gt M_2$, hence $f(x)g(x)\gt M_1M_2=M$, and thus $\lim_{x\to a}(f(x)g(x))=\infty$.

Note: These basic "arithmetic" theorems for limits do require the formal (epsilon-delta) definitions for their proofs. But once you have the theorems at your disposal, you can cut way back on the need for the formal definitions. You just have to make sure you're applying the theorems correctly.

  • $\begingroup$ The rule you stated is only valid if both limits are finite. $\endgroup$ – JLagana Mar 30 '15 at 2:02
  • $\begingroup$ @JLagana, it's true you have to be careful with limits that don't exist, but you can state a multiplication law that works when one (or, in this case, both) of the limits is infinity. I assumed you were allowing such a law. If you like, I can elaborate on what's needed in the answer. $\endgroup$ – Barry Cipra Mar 30 '15 at 13:16
  • $\begingroup$ Could you please elaborate? Can I use the basic limit laws when one or both limits don't exist? $\endgroup$ – JLagana Mar 31 '15 at 21:35

If $|x| < 1$, then $\frac1{x^2} > |\frac1{x}|$, so the $\delta$ you use to show $\frac1{x} \to \infty$ will also work for $\frac1{x^2}$.

For a proof that $\lim_{x \to a}f(x) = \infty$, you need to show that, for any $V > 0$, there is a $c(V) > 0$ such that $|x-a| < c(V)$ implies $f(x) > V$.

My comment above shows that a $c(V)$ that works for $\frac1{x}$ will works for $\frac1{x^2}$, but that is not necessary in the proof.

  • $\begingroup$ Yeah, that was my initial approach, but that resorts to the formal definition of the limit. Nothing wrong with it, but I was wondering if someone could come up with a proof that doesn't, $\endgroup$ – JLagana Mar 27 '15 at 22:04

Doing a substitution. Let $x=u^2. \\ x \rightarrow 0^+ \text{ then } u \rightarrow 0 \\\text{ So } \lim_{x \rightarrow 0^+} \frac{1}{x}=\infty \text{ implies } \lim_{u \rightarrow 0} \frac{1}{u^2}=\infty$

  • $\begingroup$ I think it could be more rigorous than that. Perhaps by splitting the $\lim_{x\to0}\frac{1}{x^2}$ in left and right ones first, then substituting. $\endgroup$ – JLagana Mar 27 '15 at 22:20
  • $\begingroup$ I'm not sure about the rigor part of it. I'm not really that great that proving things rigorously. What do you mean about splitting the thingy in left and right ones? $\endgroup$ – randomgirl Mar 27 '15 at 22:27
  • $\begingroup$ Why was this voted down? Please if there is something wrong with a substitution, let me know. $\endgroup$ – randomgirl Mar 30 '15 at 1:17
  • 1
    $\begingroup$ @randomgirl, I wasn't the one who voted you down, but I see a flaw in your reasoning: the fact that $x\to0^+$ doesn't necessarily imply that $u\to0$. $\endgroup$ – JLagana Mar 30 '15 at 2:07
  • 1
    $\begingroup$ Not the downvoter either, but what you wrote is basically already in the question. The direction of implication is also slightly fishy – to argue that $\lim_{u \to 0} 1/u^2 = \lim_{x \to 0^+} 1/x = \infty$ you want $u \to 0 \implies x \to 0^+$, not the other way around. $\endgroup$ – epimorphic Mar 30 '15 at 2:09

You could use a general result:

If there exists $\epsilon>0$ such that $f(x)\geq g(x)$ for all $0<x<\epsilon$, and $\lim_{x\to 0^+} g(x)=+\infty$, then $\lim_{x\to 0^+} f(x)=+\infty$.

Proving that lemma will require some $\epsilon-\delta$ result, but at least it separates it out into a lemma that is obviously true. It is essentially the squeeze theorem at infinity.

Alternatively, if you could use a lemma:

If $g$ is a function such that $\lim_{y\to+\infty} g(y)=+\infty$, and $f$ such that $\lim_{x\to 0^+} f(x)=+\infty$, then $\lim_{x\to 0^+} g(f(x))=+\infty$.

Basically, the claim that $\lim_{y\to+\infty} g(y) = +\infty$ is roughly stating that $g$ is continuous at $+\infty$, if $g(+\infty)$ is defined as $+\infty.$

Yet another possible lemma:

$$\lim_{x\to 0^+} f(x)=+\infty \iff \lim_{x\to 0^+} \arctan f(x) = \frac{\pi}{2}$$

This explains why my first lemma really is the squeeze theorem, since $\arctan$ is strictly increasing.

There are other functions with the same properties as $\arctan$, like $\frac{x}{\sqrt{1+x^2}}$, namely, it is strictly increasing, and:

$$\lim_{x\to 0^+} f(x)=+\infty \iff \lim_{x\to 0^+} \frac{f(x)}{\sqrt{1+f(x)^2}}= 1$$

  • $\begingroup$ Also right, but same problem as marty's. $\endgroup$ – JLagana Mar 27 '15 at 22:19
  • $\begingroup$ I've added a few other possibilities, but really, you haven't told us what theorems we are allowed to use. If we are not allowed to use any theorems, we have to revert to the definitions. @JLagana $\endgroup$ – Thomas Andrews Mar 27 '15 at 22:33
  • $\begingroup$ I stated (roughly) which theorems are allowed. Namely, the basic limit laws and the fact that $\lim_{x\to a}f(g(x)) = f(\lim_{x\to a}g(x))$. $\endgroup$ – JLagana Mar 30 '15 at 2:05
  • $\begingroup$ Sorry, but "the basic limit laws" is extremely vague. Once you enumerate what you mean, this question will be clear. In particular, the basic limit laws usually exclude the case of $\infty$. @JLagana $\endgroup$ – Thomas Andrews Mar 30 '15 at 3:21
  • $\begingroup$ Hm, you're right. The basic limits laws do exclude the case of $\infty$. $\endgroup$ – JLagana Mar 30 '15 at 3:22

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