1

I'm reading through some lecture notes to prepare myself for analysis next semester and stumbled along the following exercises:

a) Prove that $\lim_{x\to0} f(x)=b$ is equivalent to the statement $\lim_{x\to0} f(x^3)=b$.

b) Give an example of a map where $\lim_{x\to0} f(x^2)$ exists, but $\lim_{x\to0} f(x)$ does not.

for b) I was thinking about the following piecewise function:

$f(x)=\begin{cases} -1 & x < 0 \\ 1 & x \geq0 \end{cases}$

is this a good example?

for (a), I don't have any concrete tools to work with, I can't write down any explicit $\epsilon$ or $\delta$, so what can I do?

3 Answers3

1

Yes your example for point b) is a good example.

For a) the property is true for continuity of the function $x^3$ and since $x^3 \to 0$ as $x\to 0$. Yes we can prove that by the $\epsilon-\delta$ definition.

Refer also to Formal basis for variable substitution in limits.

user
  • 154,566
  • 1
    $x^2$ satisfies both condition of a) and the claim isn't true. I guess it s because $x^3$ is injective (added to the fact you say). – Dog_69 May 15 '18 at 19:35
  • interesting point, the argument would be equivalent, but wouldn't work since $x^2$ isn't injective. –  May 16 '18 at 05:34
1

For a) You can use the following in an epsilon delta proof:

for any $x \in \mathbb{R}$, $p(x) \iff$ for any $x^3 \in \mathbb{R}$, $p(x^3)$

cj1996
  • 45
  • Would this break down for even powers of p for $ f(x^p)$? –  May 16 '18 at 06:08
  • 1
    Yes, suppose you have, for any $x^p \in \mathbb{R}, p(x^p)$ is true, since $x^p \geq 0$, this does not tell us that $p(-1)$ has to be true. – cj1996 May 17 '18 at 12:56
0

Suppose, for every $\epsilon>0$, whenever $|x-0|< \delta$, it holds true that $|f(x)-b|<\epsilon$. With this in mind we can make the case for $x^3$,

By using the fact that $x^3$ is bijective and

for any $x \in \mathbb{R}$, $p(x)$ is true$\iff$ for any $x^3 \in \mathbb{R}$, $p(x^3)$ is true

since $x^3$ tends to $0$, as $x$ tends to zero, we can replace the orginal limit by $\lim_{x^3\to 0} f(x^3)$, after substituting $x $ by $x^3$ to get:

for every $\epsilon>0$, whenever $|x^3-0|< \delta$, it holds true that $|f(x^3)-b|<\epsilon$. QED.

An alternative way would be via the substitution rule for limits:

Let $\lim_{y \to 0} f(y)=0$, and notice if we choose $\lim_{x \to 0}y= \lim_{x\to 0} x^3=0$ then by substitution $\lim_{x\to 0} f(x^3)=0$,

  • The same should happen for $x^2$ but it is false. So your proof can't be valid. – Dog_69 May 16 '18 at 15:30
  • for any $x \in \mathbb{R}$, $p(x)$ is true$\iff$ for any $x^3 \in \mathbb{R}$, $p(x^3)$ is true only because $x^3$ is bijective, $x^2$ is not. –  May 16 '18 at 20:16
  • I only use one direction here, for all $x,q $ it is true that $p(x) \rightarrow p(x^q)$ the converse is not true as shown by question b) –  May 18 '18 at 06:42