Solutions

i)

Suppose

$$\lim_{x\to 0^+} f(x)=l.$$

Pick any $\varepsilon>0$. By definition, there exists $\delta>0$ such that, for all $x$,

$$0<x<\delta \Rightarrow |f(x)-l|<\varepsilon.$$

Let $x=-z$. Then,

$$0<-z<\delta \Rightarrow |f(-z)-l|<\varepsilon.$$

The proof in the other direction is similar.

ii)

Suppose

$$\lim_{x\to 0}f(|x|)=l.$$

Pick any $\varepsilon>0$. By definition, there exists $\delta>0$ such that, for all $x$,

$$0<|x|<\delta \Rightarrow |f(|x|)-l|<\varepsilon.$$

From this, we have

$$0<x<\delta \Rightarrow |f(|x|)-l|<\varepsilon \Rightarrow |f(x)-l|<\varepsilon.$$

Now, for the reverse direction $(\Leftarrow)$, suppose

$$\lim_{x\to 0^+}f(x)=l.$$

Pick $\varepsilon>0$. By definition, there exists $\delta>0$ such that

$$0<x<\delta \Rightarrow |f(x)-l|<\varepsilon.$$

Let $x=|z|$. Then,

$$0<|z|<\delta \Rightarrow |f(|z|)-l|<\varepsilon.$$

iii)

Suppose

$$\lim_{x\to 0}f(x^2)=l.$$

Pick $\varepsilon>0$. By definition, there exists $\delta>0$ such that for all $x$,

$$0<|x|<\delta \Rightarrow |f(x^2)-l|<\varepsilon.$$

Let $z=\sqrt{x}$ for $x>0$. This implies

$$0<z<\delta \Rightarrow |f(z)-l|<\varepsilon.$$

For the reverse direction $(\Leftarrow)$, suppose

$$\lim_{x\to 0^+}f(x)=l.$$

Pick $\varepsilon>0$. By definition, there exists $\delta>0$ such that, for all $x$,

$$0<x<\delta \Rightarrow |f(x)-l|<\varepsilon.$$

Let $z^2=x$. We have

$$0<x=z^2<\delta \Rightarrow 0<z<\sqrt{\delta}\ \vee\ -\sqrt{\delta}<z<0.$$

Hence,

$$0<|z|<\sqrt{\delta} \Rightarrow |f(z^2)-l|<\varepsilon.$$