Show that $\lim_{n \to \infty}\frac{d(x_{n}, x_\star)}{d(x_n, x_{n-1})} = 0$ Let $(\mathcal{X}, d)$ be a metric space and $(x_n)_{n = 0}^\infty$ a sequence in $\mathcal{X}$ converging to $x_\star \in \mathcal{X}$, meaning that $\lim_{n \to \infty}d(x_n, x_\star) = 0$
Show that
$$
\lim_{n \to \infty}\frac{d(x_{n}, x_\star)}{d(x_n, x_{n-1})} = 0
$$
(with conventions 0/0: = 0 and a/0:= $\infty$ for $a>0$)
answer:
We use the following characerisation of the limits of a ratio: 
$$
\lim_{x \to \infty}\frac{f(x)}{g(x)} =0 \iff \forall \varepsilon>0, \exists \eta >0, \forall x > \eta \quad f(x)\leq \varepsilon g(x)
$$
(This way, we don't have to distinguish when $g(x) = 0$ or not)
\newline
Let $\varepsilon>0$ and $\alpha = \frac{\varepsilon}{1 + \varepsilon} > 0$
As $\lim_{n \to \infty}\frac{d(x_{n+1}, x_\star)}{d(x_n, x_\star)} = 0$ we have:
\begin{align*}
\exists n_0, \forall n\geq n_0 - 1; \quad d(x_n, x_\star) &\leq \alpha d(x_{n- 1}, x_\star)\\
d(x_n, x_\star) &\leq \alpha \Big(d(x_{n- 1}, x_n) + d(x_{n}, x_\star)\Big)\\
d(x_n, x_\star) &\leq \frac{\alpha}{1-\alpha} d(x_{n- 1}, x_n)\\
&\leq \frac{\varepsilon / (1 + \varepsilon)}{1 - \varepsilon / (1 + \varepsilon)}d(x_{n - 1}, x_n) \qquad (1)\\
&\leq \varepsilon d(x_n, x_{n - 1}) \qquad (2)
\end{align*}
Then
$$
\frac{d(x_n, x_\star)}{d(x_n, x_{n - 1})}\xrightarrow[n \to \infty]{}0
$$
My question
How can we go from (1) to (2)?
 A: This is not true. Take $x_n=\frac 1 n$ and $x=0$ in the real line. 
A: As Kavi Rama Murthy's answer shows, the claimed limit need not be true.

As to the error in your proof attempt, it's right at the beginning, where you state


*
As$\;{\displaystyle{\lim_{n \to \infty}\frac{d(x_{n+1}, x_\star)}{d(x_n, x_\star)}}} = 0$,$\;$we have .  . .


There's no way to justify the above claim.

It's true that by hypothesis, you have
\begin{align*}
&\lim_{n \to \infty} d(x_{n+1}, x_\star)= 0\\[4pt]
&\lim_{n \to \infty} d(x_{n}, x_\star)= 0\\[4pt]
\end{align*}
but those two limits do not allow you to claim
$$\lim_{n \to \infty}\frac{d(x_{n+1}, x_\star)}{d(x_n, x_\star)} = 0$$
so the proof fails right there.

Now that I look back at your post, there's an earlier error. You assert the "convention"$\;{\large{\frac{0}{0}}}=0$,$\;$but that's not a valid convention. So perhaps that's what led you astray.

With regard to your question as to how you get from $(1)$ to $(2)$, just simplify the compound fraction:
\begin{align*}
&\frac{\varepsilon / (1 + \varepsilon)}{1 - \varepsilon / (1 + \varepsilon)}\\[4pt]
=\;&\frac{\varepsilon / (1 + \varepsilon)}{1 - \varepsilon / (1 + \varepsilon)}
\cdot \frac{1+\varepsilon}{1+\varepsilon}\\[4pt]
=\;&\frac{\varepsilon}{(1+\varepsilon)-\varepsilon}\\[4pt]
=\;&\frac{\varepsilon}{1}\\[4pt]
=\;&\varepsilon\\[4pt]
\end{align*}
