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I am trying to approximate the price of a European put with strike K and expiration T. I have to assume that the spot price $S_0$ is a lot less than $K$ or a lot more than $ K$.

I understand how this would work with digital options, where the $\frac{S_0}{K}$ would either be very large or very small. Thus, the Black-Scholes formula would yield something closer to 1 or 0.

How would I would this work with a European call or put?

My idea of a solution would be that the put price would be:

$$e^{-rT}(-S_0N(-d_1))$$

Where the $d_2$ is either ignored (if closer to 0) or just one (if closer to 1). But this seems too simplified.

What would be the correct way to approach this problem?

user_123945839432
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    Intuitively, when the strike price $K$ is way above the spot price, the call option will be worthless because no investor would be interested in buying such OTM call. It is the other way around if the spot $S(0)$ is way above the strike. This option would be very expensive since the probability to end ITM is very high. To formalise this, investigate what happens to the BS price in the extreme cases $S(0)/K \to 0$ and $S(0)/K \to \infty$. – Cavents Apr 27 '17 at 19:45

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