I am trying to solve this problem :
$$ \mathbb{P}=\lim_{x\rightarrow 0} \frac{\frac{p_{H}+(1-p_{H})(1-x)^{t}}{p_{H}+(1-p_{H})(1-x)^{t-x}}-\frac{p_{L}+(1-p_{L})(1-x)^{t}}{p_{L}+(1-p_{L})(1-x)^{t-x}}}{x} $$
where $0<p_{L}<p_{H}<1$ and $0<t<\infty$.
Both denominator and nominator goes to 0, so I thought of applying l'Hopital's rule.
As the first step, I got the following :
$$\lim_{x\rightarrow0}\frac{d\bigg{(}\frac{p_{H}+(1-p_{H})(1-x)^{t}}{p_{H}+(1-p_{H})(1-x)^{t-x}}\bigg{)}}{dx} = 0$$
using $ \lim_{x\rightarrow 0} \frac{d(1-x)^{t-x}}{dx}= -t$. The same calculation applies to the case of $p_{L}$. Therefore, I concluded that $\mathbb{P}=0$.
However, I feel something is going wrong, as $\mathbb{P}$ in my original problem represents Bayesian updating form.
Let $a=p_H$ and $b=p_L$; so the numerator of the expression is $$A=\frac{a+(1-a) (1-x)^t}{a+(1-a) (1-x)^{t-x}}-\frac{b+(1-b) (1-x)^t}{b+(1-b) (1-x)^{t-x}}$$ Using Taylor expansion around $x=0$ $$(1-x)^k=\sum_{n=0}^\infty (-1)^n \binom{k}{n} x^n$$ Use it for a very few terms and continue with the long division to obtain $$\frac{a+(1-a) (1-x)^t}{a+(1-a) (1-x)^{t-x}}=1+(a-1) x^2-\frac{1}{2} (a-1) (2 a t-1)x^3+O\left(x^4\right)$$ that is to say $$A=(a-b)x^2+O\left(x^3\right) \implies \mathbb{P}=\lim_{x\rightarrow 0} \frac A x=\lim_{x\rightarrow 0}[(a-b)x]=0 $$ which is what you found using L'Hospital.
Are you sure that the denominator is not $x^2$ ?