# What is the value of an ATM call under the Black Scholes Framework when $T \rightarrow \infty$?

In the Black Scholes framework what is the value of an at-the-money vanilla European call option as time to maturity goes to infinit ($T \rightarrow \infty$)?

• You should at the very least show what you have tried so far and where you are stuck. We are not here to do you exercise sets. – SRKX Nov 7 '16 at 6:39
• @SRKK, I am sorry if it looks like i am asking you guys for solving my questions. I didnot get any ideas how to approach this one but when Gordon posted the answer i posted my thinking as well. I will keep this in min – Raveesh Nov 7 '16 at 14:46

The Black-Scholes call option price is given by \begin{align*} C = S_0 N(d_+) - K e^{-rT}N(d_-), \end{align*} where $$d_{\pm}= \frac{\ln \frac{S_0}{K}+(r\pm\frac{1}{2}\sigma^2) T}{\sigma \sqrt{T}}.$$ Here, we assume that the interest rate $r\ge 0$ and $\sigma >0$. For an at-the-money call option, that is, $K=S_0$, we note that $\lim_{T\rightarrow \infty} d_+ = \infty$, that is, $$\lim_{T\rightarrow \infty} N(d_+) = 1.$$ In addition, if $r>0$, then $$\lim_{T\rightarrow \infty}Ke^{-rT} N(d_-) = 0,$$ and if $r =0$, then $\lim_{T\rightarrow \infty} d_- = -\infty$, and $\lim_{T\rightarrow \infty} N(d_-) = 0.$ Therefore, $$\lim_{T\rightarrow \infty} C = S_0.$$
• Pardon my ignorance but wouldnt $\lim_{T\rightarrow \infty} e^{-qt} = 0$ and thus the value of the call option be zero – Raveesh Nov 6 '16 at 18:59
• You may need to show your work. What is $q$ and how does it appear in Black-Scholes? – Gordon Nov 6 '16 at 20:43