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$X \sim \mathcal{N}(\mu,\sigma^2)$. Let $K > 0$. Prove $$\lim_{t->\infty} P(|X|>t)e^{Kt} = 0.$$

I can prove this for standard normal distribution. But I don't know how to proceed with a general normal distribution since I cannot exploit the symmetry.

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Rewrite $\mathbb{P}(|X|>t)$ as $$ \mathbb{P}(|X|>t) = \mathbb{P}(\{X > t\} \cup \{ X < - t\})=\left( 1 - \Phi\left( \frac{t-\mu}{\sigma} \right) + \Phi\left( \frac{-t-\mu}{\sigma} \right) \right), $$ and note that $\Phi\left(\frac{t-\mu}{\sigma}\right) \xrightarrow{} 1$ as $t\to \infty$ in a faster rate than $\exp\{Kt\} \to \infty$ as $\Phi$ depends on $\exp\{-(t-\mu)^2/ ( 2 \sigma^2)\}$, namely on a squared term and not linear as $\exp\{Kt\}$, hence the original limit goes to $0$.

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