Answer
$\lt 0, \dfrac{1}{2},1 \gt$
Work Step by Step
Given: $\lim\limits_{t \to \infty} \lt te^{t}, \dfrac{t^3+t}{2t^3-1}, t \sin \dfrac{1}{t} \gt$
$\lt \lim\limits_{t \to \infty} te^{t}, \lim\limits_{t \to \infty}
\dfrac{t^3+t}{2t^3-1}, \lim\limits_{t \to \infty} t \sin \dfrac{1}{t} \gt$
Need to apply L-Hospital's Rule.
$\lt \lim\limits_{t \to \infty} te^{t}, \lim\limits_{t \to \infty}
\dfrac{t^3+t}{2t^3-1}, \lim\limits_{t \to \infty} t \sin \dfrac{1}{t} \gt$
This implies that $\lt \lim\limits_{t \to \infty} \dfrac{t}{e^t}, \lim\limits_{t \to \infty}
\dfrac{1+1/t^2}{2+1/t^3}, \lim\limits_{t \to \infty} \dfrac{-1/t^2 \cos (1/t)}{-1/t^2} \gt$
This gives: $\lt 0, \dfrac{1+0}{2+0}, \lim\limits_{t \to \infty} \cos (\dfrac{1}{t}) \gt$
Answer: $\lt 0, \dfrac{1}{2},1 \gt$
