How to prove that $f(x) =1/x$ is unbounded in $(0,1)$? Let $f(x) = 1/x$ for all $x\in (0,1)$
By assuming $f(x)$ is bounded or in any other way but without using limits
I assumed that $f$ is bounded above, then there is $M>0$ s.t $f(x) \leq M$ for all $x$ in $(0,1)$.
$$1/x \leq M \implies  x \geq 1/M$$
 A: By definition, $f$ is bounded if and only if there exists $M\in\mathbb{N}$ such that for all $x\in (0,1)$, we have $|f(x)| < M$. 
Assume that such an $M$ exists, now take $x=\frac{1}{M+1}$. Then $f(x)=M+1$ and $|f(x)| > M$. Contradiction.
A: It's lower bound is obviously $1$, it is a constantly decreasing function in the range and $f(1)=1$ is defined. 
For the upper bound, assume that bound is $U=\frac{1}{x_0}$
Then evaluate $f(\frac{x_0}{2})$, we know $\frac{x_0}{2}<x_0$. Then apply the knowledge that for $a,b>0$ we have $$a>b\iff\frac ab> 1$$
What does this tell you?
A: You are on the right track. Let's assume $f(x)=1/x$ is bounded, therefore there exists $M>0$ such that $f(x)\leq M$ when $x\in(0,1)$. Without loss of generality, we can assume that $M>1$ (since if $M$ were lesser than 1, any number greater than 1 would also be a bound). 
This means that,
$1/x\leq M\implies 1\leq Mx$
The above equation is valid for any $x\in(0,1)$ and since $M>1$, $M+1>1$ and $\frac{1}{M+1}<1$. In particular $\frac{1}{M+1}\in (0,1)$, so if we pick $x=1/(M+1)$ the above inequality says that $1\leq \frac{M}{M+1}$. 
But we also know that $M<M+1$ and therefore $\frac{M}{M+1}<1$ This contradicts the above inequality. Therefore $f(x)$ is not bounded.
