Prove : $ \frac{\alpha}{\alpha -\beta} \quad \text{close to 1.}$ If $\alpha \geq |\beta^{1+\eta}|,$ how to prove that-
$$ \frac{\alpha}{\alpha -\beta} \quad \text{close to 1} ?$$
Attempt:  If,  $\alpha= |\beta^{1+\eta}|$,
then, $$ \frac{\alpha}{\alpha -\beta}= \frac{\beta^{1+\eta}}{\beta^{1+\eta} -\beta}= \frac{\beta^{\eta}}{\beta^{\eta} -1} \approx \frac{\beta^{\eta}}{\beta^{\eta} -0} \approx 1$$
but if,  $\alpha> |\beta^{1+\eta}|$ then how we extend the argument?
Note that we can not write directly $ \frac{\alpha}{\alpha -\beta}> \frac{\beta^{1+\eta}}{\beta^{1+\eta} -\beta}$ as $\alpha -\beta > \beta^{1+\eta} -\beta$
Edit:    Note, $\eta \leq 1/2$.
Source of the problem :

Full Paper: linear forms in the logarithms of real algebraic numbers close to 1, page 11.
 A: $$\frac\alpha{\alpha-\beta} = 1+\frac{\beta}{\alpha -\beta}$$
with the basic error bound
$$\left|\frac\beta{\alpha-\beta}\right| \le \frac{|\beta|}{|\beta|^{1+\eta}-s|\beta|} = \frac1{|\beta|^\eta-s} \le \frac1{|\beta|^\eta - 1}$$
where $s=\pm 1$ is the sign of $\beta$.
So if at least one of $\eta>0$, and $\beta>1$ is "large enough", then $\alpha/(\alpha+\beta)$ is "close to one".
A: Assuming $|\beta|^\eta\gt1$, then the triangle inequality guarantees
$$
\begin{align}
\left|\,\frac{\alpha}{\alpha-\beta}-1\,\right|
&=\left|\,\frac{\beta}{\alpha-\beta}\,\right|\\
&=\left|\,\frac1{\alpha/\beta-1}\,\right|\\
&\le\frac1{|\beta|^\eta-1}
\end{align}
$$
which is small if $|\beta|^\eta$ is large.
A: In the proof of $\bf{Theorem\; 13}$, $\alpha$ and $\beta$ are considered integers, large enough. 
$$\frac{\alpha}{\alpha-\beta}=\frac{1}{1-\frac{\beta}{\alpha}}\leq \frac{1}{1-\beta^{-\eta}},$$
from where the result.
I wrote only the RHS, and this just for $\beta>0.$ 
A: Since you have proven the inequality for $\alpha=|\beta^{1+\lambda}|$ notice that the function $\dfrac{\alpha}{\alpha-\beta}$ is a decreasing function of $\alpha$ when $\alpha>\beta>0$ so we have $$1<\dfrac{\alpha}{\alpha-\beta}<\dfrac{|\beta^{1+\lambda}|}{|\beta^{1+\lambda}|-\beta}\approx 1$$which means that $\dfrac{\alpha}{\alpha-\beta}$ is close to $1$ because $\dfrac{|\beta^{1+\lambda}|}{|\beta^{1+\lambda}|-\beta}$ is close to $1$
