Limit sequence proof 
Let $X=(x_n)$ be a sequence of strictly positive real numbers such
  that $\lim\left(\frac{x_{n+1}}{x_n}\right)<1$. Show that for some $r$ with
  $0<r<1$ and some $C>0$, then we have $0<x_n<Cr^n$ for all sufficently
  large $n\in \mathbb{N}$. Use this to show that $\text{lim}(x_n)=0$. Similarly, show that $\lim\left(\frac{x_{n+1}}{x_n}\right)>1$ is not bounded and hence not convergent.

Since I know that $\lim\left(\frac{x_{n+1}}{x_n}\right)<1$ and that $0<r<1$, then I have
$0<\lim\left(\frac{x_{n+1}}{x_n}\right)\le r<1$, but I get stuck now completing the proof. 
 A: You write:

Since I know that $\lim\left(\frac{x_{n+1}}{x_n}\right)<1$ and that $0<r<1$, then I have
  $0<\lim\left(\frac{x_{n+1}}{x_n}\right)\le r<1$, but I get stuck now completing the proof. 

You do know that $\lim\left(\frac{x_{n+1}}{x_n}\right)<1$, but you do not know that $0<r<1$: you are supposed to be proving that there is some real number $r$ such that $0<r<1$ (and $r$ has other properties as well).
Start by using the definition of the limit of a sequence. Let $L=\lim_{n\to\infty}\frac{x_{n+1}}{x_n}$. Then by the definition of limit you know that for each $\epsilon>0$ there is an $n_\epsilon\in\Bbb N$ such that $$\left|\frac{x_{n+1}}{x_n}-L\right|<\epsilon$$ for all $n\ge n_\epsilon$. Let $r=\frac12(L+1)$; then $L<r<1$. (Why?) Let $\epsilon=r-L$; certainly $\epsilon>0$, so we know that $$\left|\frac{x_{n+1}}{x_n}-L\right|<\epsilon\tag{1}$$ for all $n\ge n_\epsilon$. But the inequality $(1)$ is equivalent to
$$L-\epsilon<\frac{x_{n+1}}{x_n}<L+\epsilon=r\;,$$
so we now know that $\dfrac{x_{n+1}}{x_n}<r$ for all $n\ge n_\epsilon$. Let 
$$C=\frac{x_{n_\epsilon}}{r^{n_\epsilon}}\;;$$
Then $x_{n_\epsilon}=Cr^{n_\epsilon}$, and I leave it to you to show that $x_n<Cr^n$ for all $n>n_\epsilon$. Since $0<r<1$, we know that $\lim_{n\to\infty}Cr^n=0$, and since $0<x_n<Cr^n$ for all $n>n_\epsilon$, it then follows from the squeeze theorem that $\lim_{n\to\infty}x_n=0$.
A: Write the definition of $\ell=lim_{n\to+\infty}\frac{x_{n+1}}{x_n}$ and choose $\varepsilon$ such that $r=\ell+\varepsilon<1$.
