Limit of some probability of the empirical mean of independent random variables Let $X_{1}, X_{2}, \ldots, X_{n}$ be a sequence of independent, standard Normal, real-valued random variables, and consider the empirical mean $\hat{S}_{n}=\frac{1}{n} \sum_{i=1}^{n} X_{i}$. Since $\hat{S}_{n}$ is again a Normal random variable with zero mean and variance $1 / n$, it follows that for any $\delta>0$,
$$
P\left(\left|\hat{S}_{n}\right| \geq \delta\right) \underset{n \rightarrow \infty}{\longrightarrow} 0,
$$
and for any interval $A$ by CLT:
$$
P\left(\sqrt{n} \hat{S}_{n} \in A\right) \underset{n \rightarrow \infty}{\longrightarrow} \frac{1}{\sqrt{2 \pi}} \int_{A} e^{-x^{2} / 2} d x
$$
Note now that by a change of variable
$$
P\left(\left|\hat{S}_{n}\right| \geq \delta\right)=1-\frac{1}{\sqrt{2 \pi}} \int_{-\delta \sqrt{n}}^{\delta \sqrt{n}} e^{-x^{2} / 2} d x= \frac{1}{\sqrt{2 \pi}} \int_{|x|> \delta \sqrt n} e^{-x^{2} / 2} d x
$$
Now in [Large deviation techniques and applications, Amir Dembo Ofer Zeitouni] it is claimed that
$$
\frac{1}{n} \log P\left(\left|\hat{S}_{n}\right| \geq \delta\right) \underset{n \rightarrow \infty}{\longrightarrow}-\frac{\delta^{2}}{2}
$$
How do you see it?
 A: Attempting to flesh out Henry's comment:
The left-hand side is $\frac{1}{n} \log [2(1-\Phi(\delta \sqrt{n}))]$, which has the same limit as $\frac{1}{n} \log(1-\Phi(\delta \sqrt{n}))$ (if the limits exist).
From integration by parts, we have the following Mills ratio bounds for $z>0$:
$$\frac{1}{z} - \frac{1}{z^3} < \frac{1-\Phi(z)}{\phi(z)} < \frac{1}{z} - \frac{1}{z^3} + \frac{3}{z^5}.$$
[Perhaps this inequality is overkill for your particular question, is there a simpler result?]
This means
$$
-\frac{z^2}{2n} + \frac{1}{2n} \log(2\pi) + \frac{1}{n}\log\left(\frac{1}{z} - \frac{1}{z^3}\right)
< \frac{1}{n}\log(1-\Phi(z))
< -\frac{z^2}{2n} + \frac{1}{2n} \log(2\pi) + \frac{1}{n} \log\left(\frac{1}{z} - \frac{1}{z^3} + \frac{3}{z^5}\right).$$
That is,
$$
\frac{1}{n} \log(1-z^{-2})
<\frac{1}{n} \log(1-\Phi(z)) - \left(-\frac{z^2}{2n} + \frac{1}{2n} \log(2\pi) - \frac{\log z}{n}\right)
< \frac{1}{n} \log(1-z^{-2} + 3 z^{-4})$$
If you plug in $z=\delta\sqrt{n}$ you can show $-\frac{z^2}{2n} + \frac{1}{2n} \log(2\pi) - \frac{\log z}{n} \to -\frac{\delta^2}{2}$ and that the two outer bounds converge to zero.
A: While the existing answer is nice and absolutely correct, it still misses important general techniques from the large deviation theory, so I will add mine.
Concerning the upper bound, it is usually obtained with the help of Chernoff inequality:
$$
\mathrm P(\hat S_n\ge \delta) \le \frac{\mathrm E[e^{\lambda \hat S_n}]}{e^{\lambda \delta}} = \exp\Big\{\frac{\lambda^2}{2n} - \lambda \delta\Big\}, \lambda >0,
$$
whence, choosing $\lambda = n\delta$ (which minimizes the left-hand side),
$$
\mathrm P(\hat S_n\ge \delta) \le e^{-n\delta^2/2}.
$$
Similarly, $\mathrm P(\hat S_n\le -\delta) \le e^{-n\delta^2/2}$,
so
$$
\limsup_{n\to\infty} \frac{1}{n}\log \mathrm P(|\hat S_n|\ge \delta)\le \limsup_{n\to\infty} \frac{1}{n}\Big( -\frac{n\delta^2}2 +\ln 2\Big) = -\frac{\delta^2}2. 
$$
To show the lower bound, it is often enough to localize the integrand "near the boundary":
$$
\mathrm P(|\hat S_n|\ge \delta) = \frac{1}{\sqrt{2 \pi}} \int_{|x|\ge \delta \sqrt n} e^{-x^{2} / 2} d x\ge \frac{1}{\sqrt{2 \pi}} \int_{\delta \sqrt n}^{\delta\sqrt{n}+1} e^{-x^{2} / 2} d x \ge \frac{1}{\sqrt{2 \pi}}\cdot \exp\Big\{-\frac{(\delta\sqrt{n}+1)^2}{2}\Big\},
$$
whence
$$
\liminf_{n\to\infty} \frac{1}{n}\log \mathrm P(|\hat S_n|\ge \delta)\ge \liminf_{n\to\infty} \frac{1}{n} \Big(-\frac{(\delta\sqrt{n}+1)^2}{2}-\ln \sqrt{2\pi}\Big) = -\frac{\delta^2}{2}.
$$
