Solving $\lim_{t \to \infty} t\log(\frac{\frac{\log(\alpha + 1)}{t} - \frac{\log(t + \alpha)}{t}}{ 1 - \frac{1}{t(t + \alpha)}} + 1)$

Question

Given the following problem

$$\begin{equation} \begin{split} {\label{limit}} \lim_{t \to \infty} t\log\left(\dfrac{\dfrac{\log(\alpha + 1)}{t} - \dfrac{\log(t + \alpha)}{t}}{ 1 - \dfrac{1}{t(t + \alpha)}} + 1\right) \end{split} \end{equation}$$

where $t \in \mathbb{N}$, and $\alpha > 0$ how can I compute such a limit?

Edit: I had forgetten the + 1 inside the log in the original question. I added it.

Answer 1

(...new answer because the question was edited to yield a different result)

Assuming $\log=\ln$, it's $$L=\lim_{t\to\infty} t \ln a_t$$ $\def\a{\alpha}$ where $$\begin{align} a_t &= 1+\frac{\dfrac{\ln(\alpha + 1)}{t} - \dfrac{\ln(t + \alpha)}{t}}{ 1 - \dfrac{1}{t(t + \alpha)}} \\ &= 1-\frac{\ln\dfrac{t+\a}{\a+1}}{t-\dfrac{1}{t+\alpha}} =:1-x(t)\\ \end{align}$$ Hence $a_t\to 1$ and we can use the expansion of $\ln$ at 1 which is:

$$\ln(1-x)=-x+\Theta(x^2) \quad\text{ for }|x|<1$$

provided $t$ is large enough, and where $\Theta$ denotes a Landau symbol:

$$\begin{align} t\ln a_t &= t\ln(1-x) \\ &= -t(x+\Theta(x^2)) \\ &= -\frac{\ln\dfrac{t+\a}{\a+1}}{1-\dfrac{1}{t(t+\a)}} + \Theta (tx^2)\\ &\to-\infty \end{align}$$ and where the error term is of order $(\ln^2t)/t$.