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> If $X$ is Standard Normal and $a>0$ is a constant then find $\displaystyle\lim\_{x\to0}\mathbb{P}\left(X>x+\dfrac{a}{x}\Big|X>x\right)$.
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This is an exercise from a book whose name I cannot immediately recall. I compiled problems from different books and am stuck at one of these. The answer to this is $\mathrm{e}^{-a}$.

However, I have some doubts regarding the validity of this answer. If $x\to0$, consider the negative subsequence of this sequence, and then $$\mathbb{P}\left(X>x+\dfrac{a}x\Big|X>x\right)=\dfrac{\mathbb{P}\left(X>x+\dfrac{a}x,X>x\right)}{\mathbb{P}(X>x)}=\dfrac{\mathbb{P}(X>x)}{\mathbb{P}(X>x)}=1,$$
so this subsequence always goes to $1$, hence the limit can't be $\mathrm{e}^{-a}$ which renders the question meaningless.

However, supposing I restrict the question to finding the limit as $x\to0^+$, then I get $$\dfrac{\mathbb{P}\left(X>x+\dfrac{a}x,X>x\right)}{\mathbb{P}(X>x)}=\dfrac{\mathbb{P}\left(X>x+\dfrac{a}x\right)}{\mathbb{P}(X>x)}$$.

Now as $x\to0^+$ the limit of this quantity is $\dfrac{\mathbb{P}\left[X>\displaystyle\lim\_{x\to0}\left(x+\dfrac{a}x\right)\right]}{\mathbb{P}\left(X>\displaystyle\lim\_{x\to0}x\right)}=0$ by continuity property of probability.

Hence I am not getting $\mathrm{e}^{-a}$. Whenever $x\to0$, the positive subsequence yields answer as $0$ and the negative subsequence yields answer as $1$.

Any help will be appreciated.