Answer
(i)
$\lim_{x \to 0}f(x)=0$ does not exist.
(ii)$$\lim_{x \to 0} g(x) =0$$
Work Step by Step
(i)
We claim that the function $f(x)$ does not have limit at $x=0$.
We prove the claim by contradiction. Suppose that the function $f(x)$ has limit $L$ at $x=0$. So for $\varepsilon =\frac{1}{2}$ there must exist a $\delta _0$ such that$$|x-0|< \delta _0 \quad \Rightarrow \quad |f(x)-L| < \frac{1}{2}.$$As we know, between any two real numbers there exist a rational and irrational numbers, so in any interval there exist a rational and irrational numbers within the interval. So let us choose some rational number $y$ and irrational number $z$ within the interval $(0 - \delta _0 , 0+\delta _0 )$. According to the definition the function $f(x)$, we can get the following by substituting them into the above inequality:$$|f(y)-L|=|0-L|=|L| 0$, there exists a $\delta >0$ such that $|g(x)-0|< \varepsilon$ whenever $|x-0|< \delta$.
For any $\varepsilon >0$ we choose $\delta = \varepsilon$. Let $x$ be any real number within the interval $(0- \delta , 0+\delta )$. As we know, between any two real numbers there exist rational and irrational numbers. So there are two cases: (a) $x$ is rational; in this case we have $g(x)=0$, according to the definition of $g(x)$, so trivially the inequality $|g(x)-0| 0$, there exists a $\delta >0$ such that$$|x-0|
