Prove a conclusion of LDP: $\frac{1}{n}\log P(|\hat{S}_n|\geq\delta)\xrightarrow{n\xrightarrow{}\infty}-\frac{\delta^2}{2}$

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I plan to learn something about Large Deviation Principle (LDP). I am reading the book Large Deviations Techniques and Applications and immediately confused by a problem in the first page.

The original text is

Note that

$P(|\hat{S}_n|\geq\delta)=1-\frac{1}{\sqrt{2\pi}}\int_{-\delta\sqrt{n}}^{\delta\sqrt{n}}e^{-x^2/2}dx$

therefore,

$\frac{1}{n}\log P(|\hat{S}_n|\geq\delta)\xrightarrow{n\xrightarrow{}\infty}-\frac{\delta^2}{2}$

I have no idea how the results is gotten. The problem may be concluded as

Proof: For any $\delta$, $\frac{1}{n}\log\left(1-\frac{1}{\sqrt{2\pi}}\int_{-\delta\sqrt{n}}^{\delta\sqrt{n}}e^{-x^2/2}dx\right)\xrightarrow{n\xrightarrow{}\infty}-\frac{\delta^2}{2}$