1. **State the problem:** Find the limit as $t \to +\infty$ of $$\frac{2}{t^2} - 4t$$ using limit theorems. 2. **Recall theorem 13 (Sum/Difference of limits):** If $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$, then $$\lim_{x \to a} [f(x) \pm g(x)] = L \pm M$$ provided both limits exist. 3. **Apply the theorem:** Let $f(t) = \frac{2}{t^2}$ and $g(t) = 4t$. 4. **Find each limit separately:** - $$\lim_{t \to +\infty} \frac{2}{t^2} = 0$$ because as $t$ grows large, $t^2$ grows large and the fraction tends to zero. - $$\lim_{t \to +\infty} 4t = +\infty$$ since $4t$ grows without bound. 5. **Combine limits:** $$\lim_{t \to +\infty} \left( \frac{2}{t^2} - 4t \right) = 0 - (+\infty) = -\infty$$ 6. **Interpretation:** The expression tends to negative infinity as $t$ approaches positive infinity. **Final answer:** $$\boxed{-\infty}$$