Spivak, Ch 5 Limits, Problems 10c: Proving limit relationship

[zenterix]

Homework Statement

10) c) Prove that $\lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} f(x^3)$ (First see why the assertion is obvious; then provide a rigorous proof)

Relevant Equations

I suppose this problem isn't super hard. I find it tricky to explain the solution in words and then in symbols, in a way that is completely rigorous and doesn't assume any steps are so obvious that they can be skipped. My goal with the solution below is to try to actually prove the relationship in a way that is correct, precise, and doesn't leave absolutely anything implicit.

c) Why is the assertion $\lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} f(x^3)$ obvious?

First of all I don't think it is obvious but here is an explanation of why the limits are the same.

$\lim\limits_{x\to0} f(x^3)=l_2$ means we are looking at points with $x$ close to zero and evaluating the function $f(x^3)$ at such points, checking if it stays close to some value $l_2$.

Whereas in $\lim\limits_{x \to 0}f(x)$ we are evaluating $f$ at points $x$ close to zero, in $\lim\limits_{x\to0} f(x^3)$ we are evaluating $f$ at points $x^3$ with $x$ close to zero.

Keeping $x$ a distance $\delta_1=\sqrt[3]{\delta}$ from zero but then evaluating $f$ at points $x^3$ is the same as keeping $x$ a distance $\delta$ from zero, but then evaluating $f$ at points $x$.

Therefore, if $|f(x)-l|<\epsilon$ when $|x-0|<\delta$ then $|f(x^3)-l|<\epsilon$ when $|x^3-0|<\sqrt[3]{\delta}$, which we can write as $\lim\limits_{y \to 0} f(y)=l$, for $y=x^3$.

Proof

$\lim\limits_{x \to 0}f(x)=l$ means that $\forall \epsilon>0$, $\exists \delta>0$ such that $|x-0|<\delta \implies |f(x)-l|<\epsilon$

$\lim\limits_{x \to 0} f(x^3)=l_2$ means that $\forall \epsilon>0$, $\exists \delta_1>0$ such that $|x-0|<\delta_1 \implies |f(x^3)-l_2|<\epsilon$.

Let $y=x^3 \implies x = \sqrt[3]{y}$

$|\sqrt[3]{y}|<\delta_1 \implies |y|<\delta_1^3=\delta_2 \implies |f(y)-l_2|<\epsilon$

$\implies \lim\limits_{y \to 0} f(y) = l_2$

$\implies l_2 = l_1$

Notice that $\delta_1^3=\delta \implies \delta_1 = \sqrt[3]{\delta}$

If $0<\delta<1$ then $\delta_1>\delta$, if $0<1<\delta$ then $\delta_1<\delta$. This is interesting but is not important in the proof.

 

[PeroK]

I'm too tired to look at your proof, but I wanted to make the observation that $x \rightarrow x^3$ is a continuous bijection from $\mathbb R$ to itself.

I wouldn't say that that was obvious. For example if we replace $x^3$ by $x^2$ here then the result does not necessarily hold.

Also, what about replacing $x^3$ with $\sin x$?

 

[PeroK]

I guess the other issue is whether we can assume that the limits both exist?

 

[zenterix]

x→x3 is a continuous bijection from R to itself.

Perhaps not obvious but also not not obvious. I am only using concepts from the current and previous chapter I am on in Spivak, so no bijections yet.

Also not inverse functions, which is what I think might be needed to consider the case of replacing $x^3$ with $\sin{x}$. However, using inverse sin, I think the proof is similar to the $x^3$ case:

$$\lim\limits_{x \to 0} f(\sin{x})=l$$ means

$\forall \epsilon>0, \exists \delta>0: |x|<\delta \implies |f(\sin{x})-l|<\epsilon$

Let $y=\sin{x}$. Then $x=\sin^{-1}{y}$, $y \in [-1,1]$.

$$(|x|<\delta \implies |\sin^{-1}{y}|<\delta \implies |y|<\sin{\delta}) \implies |f(y)-l|<\epsilon$$

$$\implies \lim\limits_{y \to 0} f(y) = l$$

The gist of these proofs is:

Assuming $\lim\limits_{x \to 0} f(\sin{x})$ exists, then keeping $x$ within $\delta$ of $0$ means $f(\sin{x})$ is within $\epsilon$ of $l$.

But if $x$ is actually a function $x=\sin^{-1}{y}$ of $y$, then we can determine a $\delta_1$ such that if $y$ is within $\delta_1=\sin{\delta}$ of $0$, then $x$ is within $\delta$ of $0$, and so $f(\sin{x})=f(y)$ is within $\epsilon$ of $l$.

I guess the other issue is whether we can assume that the limits both exist?

In this proof, I indeed assumed that both limits exist, in particular I assumed $\lim\limits_{x \to 0} f(\sin{x})$ exists.

Of course if the second limit in $\lim\limits_{x \to 0} f(x)=\lim\limits_{x \to 0} f(\sin{x})$ doesn't exist then the statement is false, so we can't prove that it is true.

If we had instead replaced $x^3$ with $\frac{1}{x}$ we would have had to assume that $\lim\limits_{x \to 0} f(\frac{1}{x})$ exists. An interesting question is, for what functions does it actually exist? Clearly not all of them: $f(x)=x \implies f(\frac{1}{x})=\frac{1}{x}$, which does not have a defined limit at $x=0$.

As an example of a function where the limit does exist, consider $f(x)=\frac{1}{1+x}$. $f(\frac{1}{x})=\frac{1}{\frac{1}{x}+1}=\frac{x}{1+x}$ and $\lim\limits_{x \to 0} f(\frac{1}{x}) = 0$.

If we assume we are dealing with one of the $f(\frac{1}{x})$ for which the limit does exist, we have:

$$\lim\limits_{x \to 0} f(\frac{1}{x})=l_2$$

means

$\forall \epsilon > 0 \exists \delta>0 : |x|<\delta \implies |f(\frac{1}{x})-l_2|<\epsilon$

Let $y=\frac{1}{x}$. Then $x=\frac{1}{y}$.

$$(|x|<\delta \implies |\frac{1}{y}|<\delta \implies |y|>\frac{1}{\delta}) \implies |f(y)-l_2|<\epsilon$$

$$\implies \lim\limits_{y \to \infty}f(y) = l_2$$

However, I am not sure how to reconcile this with

$$\lim\limits_{x \to 0} f(x)=l$$

Such that I can conclude that $l=l_2$. It is clear intuitively, but I mean notation-wise and in terms of the definitions, how do I reconcile the two?

 

[PeroK]

Proof

$\lim\limits_{x \to 0}f(x)=l$ means that $\forall \epsilon>0$, $\exists \delta>0$ such that $|x-0|<\delta \implies |f(x)-l|<\epsilon$

$\lim\limits_{x \to 0} f(x^3)=l_2$ means that $\forall \epsilon>0$, $\exists \delta_1>0$ such that $|x-0|<\delta_1 \implies |f(x^3)-l_2|<\epsilon$.

Let $y=x^3 \implies x = \sqrt[3]{y}$

$|\sqrt[3]{y}|<\delta_1 \implies |y|<\delta_1^3=\delta_2 \implies |f(y)-l_2|<\epsilon$

$\implies \lim\limits_{y \to 0} f(y) = l_2$

$\implies l_2 = l_1$

Notice that $\delta_1^3=\delta \implies \delta_1 = \sqrt[3]{\delta}$

If $0<\delta<1$ then $\delta_1>\delta$, if $0<1<\delta$ then $\delta_1<\delta$. This is interesting but is not important in the proof.

I'm not convinced by the logic here. You're still overusing $\Rightarrow$ to connect the steps in your proof. Rather than more precisely to say when one statement directly implies another.

Also, I'm not convinced you are manipulating the estimates correctly. I got confused by your $\delta, \delta_1. \delta_2$.

Finally, what you ought to be proving is something like:

a) If $\lim_{x \rightarrow 0} f(x) = l$, then $\lim_{x \rightarrow 0} f(x^3) = l$, where $l \in \mathbb R$. And, the converse.

b) If $\lim_{x \rightarrow 0} f(x) = \pm \infty$, then $\lim_{x \rightarrow 0} f(x^3) = \pm \infty$. And the converse.

The way I would do a) is as follows:

Assume $\lim_{x \rightarrow 0} f(x) = l$, where $l \in \mathbb R$.

Let $\epsilon > 0$ (note that I start this way and not with the definition of the limit!).

Then we can find $\delta > 0$ such that $|x| < \delta \ \Rightarrow \ |f(x) - l| < \epsilon$

Note that $\delta^{1/3} > 0$ and $|x| < \delta^{1/3} \Rightarrow |x^3| < \delta \Rightarrow |f(x^3) - l| < \epsilon$.

As $\epsilon$ was arbitrary, we have $\lim_{x \rightarrow 0} f(x^3) = l$.

Proving the converse is similar. And the proof of b) for the unbounded case is much the same.

 

[PeroK]

PS I would have the same approach for the $\sin x$ case, but you need some sort of relationship between $x$ and $\sin x$ for "small" $x$. E.g. you could prove that $\exists \delta_0 > 0$, such that:
$$|x| < \delta_0 \Rightarrow \frac 1 2|x| < \sin x < |x|$$

Or, you could use the fact that $\sin x$ is continuous at $x = 0$.