If $X$ is standard Normal then find $\lim_{x\to0}P(X>x+\frac{a}{x}|X>x)$ 
If $X$ is Standard Normal  and $a>0$ is a constant then find $\displaystyle\lim_{x\to0}\mathbb{P}\left(X>x+\dfrac{a}{x}\Big|X>x\right)$.

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.

I agree with your analysis: the limit does not exist, since the right- and left-sided limits differ.
I suspect that either the original problem was erroneous, or perhaps you made a mistake when copying it.
