Real analysis elementary limit property 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?
 A: 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.
A: For a) You can use the following in an epsilon delta proof:
for any $x \in \mathbb{R}$, $p(x) \iff$ for any $x^3 \in \mathbb{R}$, $p(x^3)$
A: 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   $\lim_{x\to 0} f(x^3)=0$,
