Proof that $\lim_{x\to\infty} b^x=0 \iff 0 \leq b<1$

Question

Are there any errors in the following attempt to prove the above?

$(\Leftarrow)$ Let $f(x)=b^x$, with $0 \leq b<1$. Then, for all $x$, $f(x)>0$ and $f'(x)=b^x \ln(b)<0$. This means that $f$ is decreasing and bounded below by 0, and so $\lim_{x\to\infty} f(x)=0$.

($\Rightarrow)$ Let $f(x)=b^x$. In order for $f$ to be defined, $b \geq 0$. If $\lim_{x\to\infty} f(x)=0$, then for any $\epsilon>0$ there is an $N$ such that, for all $x$, $x>N \implies |f(x)-0|=f(x)<\epsilon$. This can be the case only if $0 \leq b<1$.

Answer 1

Your $\implies$ is not solid. I think that it is better to do it by contraposition. If $b \ge 1$, then $b^x = \exp(x \log b) \ge \exp(0) = 1$ for all $x \ge 0$. So $\lim_{x \to \infty} f(x) \neq 0$.

Answer 2

A different thought on this, probably not as rigorous as Ahmed's. If $0<b<1$ then we can write $b=\frac{p}{q}$ where $q>p$. There exist a (fractional) value of $r>0$ such that $q=pr$ so that we can write $b=\frac{p}{pr}=\frac{1}{r}$ Now consider $(\frac{1}{r})^x=\frac{1}{r^x}$ when $x$ goes to infinity