Prove that $f(x)=\frac{1}{x}$, $g(x)=x$ are unbounded on $(0,\infty)$.
We show that $f$ is unbounded by assuming that there is a bound $M>0$ and then arriving at a contradiction. From the graph of $f(x)=\frac{1}{x}$, it is clear that $f(x)$ is large for values of $x$ which are close to zero. Choose $x>0$ such that $x<\frac{1}{M}$. Then $f(x)=\frac{1}{x}>M$, and so $M$ is not a bound for the function. Therefore $f$ has no bound and so is an unbounded function.
I proved the function $f(x)$. The function $g(x)$ also wants to prove similar to that of $f(x)$. But I don't know what to do.
Write the definition of boundedness neatly.
Suppose $g$ is bounded on $(0,\infty)$.
Then, there exists $M>0$ s.t. $\bigg[ |g(x)|<M$ for all $x\in (0,\infty) \bigg]$.
Since $|g(x)|<M$ holds for all $x\in (0,\infty)$, $|g(x)|<M$ must hold for $x=M+1\in (0,\infty).$
Then, $|g(M+1)|<M$. Can you lead a contradiction ?