Find $\lim_{t \to \infty} P (X>t+ \frac{x}{t} | X > t)$ for $X$ ~ $N(0,1)$

Question

Let $X$ ~ $N(0,1)$ be a random variable with normal distribution.

Find $\lim_{t \to \infty} P (X>t+ \frac{x}{t} | X > t)$

Attempt

$P (X>t+ \frac{x}{t} | X > t) = \frac{P (X>t+ \frac{x}{t})}{P (X>t)} = \frac{1 - P (t < X <t+ \frac{x}{t})}{P (X>t)}$

It 'seems' that the numerator converges to 1 while the denominator goes to $0$. I've run out of ideas. A hint would be much appreciated

Answer 1

$$P (X>t+ \frac{x}{t} | X > t) = \frac{P (X>t+ \frac{x}{t})}{P (X>t)} =\dfrac{Q(t+\dfrac{x}{t})}{Q(t)}$$where $Q(.)$ is the famous $Q$ function. Since the both the numerator and denominator tend to zero we use L'Hopital's rule and get $$\lim_{t\to\infty}\dfrac{Q(t+\dfrac{x}{t})}{Q(t)}=\lim_{t\to\infty}\dfrac{(1-\dfrac{x}{t^2})\dfrac{1}{\sqrt{2\pi}}e^{-\dfrac{(t+\dfrac{x}{t})^2}{2}}}{\dfrac{1}{\sqrt{2\pi}}e^{-\dfrac{t^2}{2}}}=\lim_{t\to \infty}\dfrac{e^{-\dfrac{1}{2}(t^2+2x+\dfrac{x^2}{t^2})}}{e^{-\dfrac{t^2}{2}}}=e^{-x}$$