Calculate $\lim_{x \to \pm \infty} \ln(\frac{x}{(x+1)^{\\\alpha}}) + \beta x$ How can I evaluate the following limit? 
$$\lim_{x \to \pm \infty} \ln(\frac{x}{(x+1)^{\\\alpha}}) + \beta x,$$
for $\alpha, \beta \in \mathbb{R}$.
 A: With equivalents at $+\infty$:
$$ f(x)=\ln\frac{x}{(x+1)^{\\\alpha}} + \beta x=\ln x -\alpha\ln(x+1) +\beta x$$


*

*If $\beta\neq 0$, it is easy to check that $\ln x -\alpha\ln(x+1)=_{+\infty}o(\beta x)$, hence
$$f(x)=\ln x -\alpha\ln(x+1) +\beta x \sim_{+\infty} \beta x$$
and $$\lim_{x\to+\infty}f(x)=\lim_{x\to+\infty}\beta x=\begin{cases}+\infty&\text{if}\enspace\beta >0\\-\infty&\text{if}\enspace\beta <0\end{cases}$$

*If  $\beta\neq 0$, $f(x)=\ln x -\alpha\ln(x+1)=(1-\alpha)\ln x-\alpha\ln\Bigl(1+\dfrac1x\Bigr) $. The second term has limit $0$ at $+\infty$, hence
$$\lim_{x\to+\infty}f(x)=\lim_{x\to+\infty}(1-\alpha)\ln x=\begin{cases}+\infty&\text{if}\enspace\alpha <1\\-\infty&\text{if}\enspace\alpha>1\\0&\text{if}\enspace\alpha=1\end{cases}$$


At $-\infty$, it is more delicate, as $f(x)$ is defined for $x<0$ only if $\alpha$ is an odd integer. In that case, setting $x=-y\enspace(y>0)$ shows we have, mutatis mutandis, the same conclusion.
A: I would say the only thing that might help is to notice that $\beta x = \textrm{ln}(e^{\beta x})$, so that you could then combine the limit into a single expression, $$\lim_{x \to \pm\infty}\ln(\frac{xe^{\beta x}}{(x+1)^\alpha}),$$ and from there, it's a matter of cases for alpha and beta.
