It is maybe a simple question but right now I am not able to see it.
For $r,q,B>0$ and $x \in \mathbb{R}$, why is the following limit equal to $1$: $$\lim_{d\to 0^+}\exp\left[\left(\frac{d}{1-q}\right)\log\left(\log B+\frac1d\log\left(1+\frac{x}{rq}\right)\right)\right]=1$$
I am thinking it has to do with an expansion of $\log\left( \log (x)\right)$?
Thank you!
Well, $$ \lim_{d \to 0^+}\ \exp\left[\left({d \over 1-q}\right)\log\left(\log B + {1 \over d}\log \left( 1+{x \over rq} \right) \right) \right]$$ equals $$ \lim_{d \to 0^+}\ \exp\left[\log\left(\log B + {1 \over d}\log \left( 1+{x \over rq} \right) \right) \right]^{{d \over 1-q}} $$ equals $$ \lim_{d \to 0^+}\ \left(\log B + {1 \over d}\log \left( 1+{x \over rq} \right) \right)^{{d \over 1-q}} $$ and take the limit from there.
More: Since $B$, $x$, $r$, and $q$ are constants, the limit has the form $$ \lim_{d \to 0^+}\ \big(a+{b \over d}\big)^{d} $$ which equals $$ \lim_{d \to 0^+}\ \big({ad+b \over d}\big)^{d} $$ which equals $$ \lim_{d \to 0^+}\ {(ad+b)^b \over d^b} $$ Now the limit $$ \lim_{x \to 0} {(ax+b)^n \over x^n}, \qquad n \in \mathbf{R}\backslash\{0\} $$ can be shown to be $a$ by L'Hopital's rule. Choose some $n$, and the limit we are working with becomes $$ \lim_{d \to 0^+}\ \left({(ad+b)^n \over d^n}\right)^{d \over n} $$ Where the limit of the inside is simply $a$, so overall the limit tends to $1$.
Advice: you have too many symbols in the expression that serve no purpose in computing the limit, and just make things look unnecessarily complicated. Get "rid" of them.
Since $r,q,B$ and $x$ are simply nice constants and $d$ is the only parameter that varies, we can rewrite $$ \exp\left[\left(\frac{d}{1-q}\right)\log\left(\log B+\frac1d\log\left(1+\frac{x}{rq}\right)\right)\right] $$ as the simpler-looking $$ \exp\left[\alpha d\log\left(\beta+\frac{\gamma}{d}\right)\right] $$ for some (related) constants $\alpha,\beta,\gamma$. To prove this converges to $1$ when $d\to 0^+$ is equivalent, by continuity of the exponential, to proving the exponent converges to $0$. From there, observe that $$ d\log\left(\beta+\frac{\gamma}{d}\right) = d\log\left(\frac{1}{d}\left(\gamma+d\beta\right)\right) = d\log\frac{1}{d} + d\log\left(1+d\frac{\beta}{\gamma}\right) $$ and both terms converge to $0$ when $d\to 0^+$ (the first as $\lim_{u\to 0^+} u\log u = 0$, the second as the product of $d\to 0$ and $\log\left(1+d\frac{\beta}{\gamma}\right) \to \log(1+0)=0$). Multiplying by the constant $\alpha$ does not change anything, so the exponent does converge to $0$ as claimed.
Sketch: This is the same as saying the expression inside the brackets $\to 0.$ Inside the brackets you have, essentially, $d\log (C_1 +C_2/d).$ That indeed goes to $0.$
