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The first limit in the PDF has a reciprocal error when $\lambda$ denotes point density. The correct strong law for Poisson arrival times isTo prove it, put $N(t)=\Pi ((0,t])$. The unit-interval increments $N(j)-N(j-1)$ are independent and identically distributed random variables with the Poisson distribution of mean $\lambda$. The strong law of large numbers gives $N(m)/m\to \lambda$ along integers. For $m\le t<m+1$,so $N(t)/t\to \lambda$ for all real $t\to \infty$. The continuous intensity gives distinct points almost surely, and local finiteness and infinitely many points give $Y_n\to \infty$. Since $N(Y_n)=n$, the same pathwise limit along $Y_n$ implies $n/Y_n\to \lambda$, proving the reciprocal limit. For instance at density $2$ the limit is $1/2$, not $2$. The printed limit would instead describe a parameter interpreted as mean interarrival time, but not the stated density.

For the Non-homogeneous Poisson point process on $(0,1)$, every closed subinterval strictly inside $(0,1)$ has finite intensity. The intensity is non-atomic, so the process is simple and has no point at $1/2$ almost surely. Both half-intervals have infinite intensity: the density behaves as $x^{-2}$ at zero and as $(1-x{)}^{-1}$ at one. More explicitly, exhaust either half by finite-intensity intervals of means $m_j\to \infty$. For any fixed integer $K$,Thus there are infinitely many points on each side.

Local finiteness around $1/2$ gives a last point on its left and a first point on its right. To see the former directly, choose any left point; between it and $1/2$ there are only finitely many points, so there is a largest one. Repeat below each chosen point to label the left sequence by $X_{-1},X_{-2},\dots$, and similarly label the right sequence by $X_0,X_1,\dots$. No infinite sequence of points can accumulate in the interior, because a compact neighbourhood there has finite count. ConsequentlyThis constructs the required two-sided order statistics of a boundary-singular Poisson process.

Use its integrated intensity to find the rates. Partial fractions giveThe left cumulative-intensity coordinate isIt maps the left half bijectively onto $(0,\infty )$, reversing its order. The image points form a unit-rate Poisson process directly from the count definition: inverse images of disjoint parameter intervals are disjoint, and their intensity equals their parameter lengths. In particular $L(X_{-k})$ is its $k$th arrival, so the result just proved gives $L(X_{-k})/k\to 1$. Also $xL(x)\to 1$ as $x\downarrow 0$, because $x\log x\to 0$. HenceThis is exactly the requested negative-index limit.

For the right side, the increasing coordinate isIts image is another unit-rate Poisson process; $F(X_n)$ is the $(n+1)$st arrival because the first right point has index zero. Thus $F(X_n)/(n+1)\to 1$. As $x\uparrow 1$,Combining these facts gives the positive-index asymptoticsEquivalently $1-X_n=\exp (-n+o(n))$ almost surely. In particular $n^p(1-X_n)\to 0$ for every fixed $p>0$, unlike the inverse-linear scale at the left endpoint.

There is also an exact distributional description, if desired. For $x\in (1/2,1)$, $X_n>x$ means that at most $n$ right-hand points precede $x$, soThus $F(X_n)$ has the gamma distribution with shape $n+1$ and rate one. This makes the indexing shift explicit and describes the fluctuations left unspecified by the logarithmic limit.