Answer
Please see below.
Work Step by Step
(i)
We first show that the function $f(x)$ is continuous at $x=0$.
We want to prove that $$\lim_{x \to 0}f(x)=0$$ by using $\varepsilon - \delta$ definition; that is, we must show that for each $\varepsilon >0$, there exists a $\delta >0$ such that $|f(x)-0|< \varepsilon$ whenever $|x-0|< \delta$.
For any $\varepsilon >0$ we choose $\delta = \frac{\varepsilon }{|k|}>0$ (Please note that $k \neq 0$). 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 $f(x)=0$, according to the definition of $f(x)$, so trivially the inequality $|f(x)-0| 0$, there exists a $\delta >0$ such that$$|x-0|0$, there exists a $\delta >0$ such that$$|x-c|< \delta \quad \Rightarrow \quad |f(x)-L| < \varepsilon .$$By choosing $\varepsilon = \varepsilon _0 = \frac{|c| |k|}{4}>0$ (Please note that $c \neq 0$ and $k \neq 0$), there exists a $\delta _1$ such that$$|x-c|< \delta _1 \quad \Rightarrow \quad |f(x)-L| < \varepsilon _0 .$$ Since $\delta _0=\min \{ \delta _1 , \frac{|c|}{2} \} < \delta _1$, we can get$$|x-c|< \delta _0 \quad \Rightarrow \quad |f(x)-L| < \varepsilon _0 .$$As we know, between any two real numbers there exist a rational and irrational numbers, so in any interval there exist rational and irrational numbers within the interval. So let us choose some rational number $y$ and irrational number $z$ within the interval $(c - \delta _0 , c+\delta _0 )$. According to the definition of the function $f(x)$, we can get the following by substituting them into the above inequality:$$|f(y)-L|=|0-L|=|L| \frac{|c|}{2}$.
Thus, $f(x)$ does not have limit at any $x \neq 0$.
