Limit of a infimum

Question

I'm doing some numerical analysis and I realized that

$$\lim_{t \to \infty} \left [\inf_{y \in \mathbb{R}} \left \{ \frac{y^2}{2t}+\sin{(y)} \right \} \right ]=-1.$$

So I must prove that $\frac{y^2}{2t} \to 0$ as $t \to \infty$ but can't find the correct argument.

Answer 1

Hint. Note that for $t>0$, $$0-1=\inf_{y\in\Bbb R} \left( \frac{y^2}{2t}\right)+\inf_{y\in\Bbb R} \left( \sin y\right) \le\ \inf_{y\in\Bbb R} \left( \frac{y^2}{2t} +\sin y\right) \ \le\ \inf_{y\in \{-\pi/2\}} \left( \frac{y^2}{2t} +\sin y\right)= \frac{\pi^2}{8t} -1.$$

Answer 2

For a fixed $t>0$, we have $$-1\ \le\ \inf_{y\in\Bbb R} \left( \frac{y^2}{2t} +\sin y\right) \ \le\ \frac{(-\pi/2)^2}{2t} - 1$$ and it tends to $-1$ if $t\to\infty$.