Answer
$$\lim_{t\to-\infty}e^{3t}\sin^{-1}\frac{1}{t}=0$$
Work Step by Step
$$A=\lim_{t\to-\infty}e^{3t}\sin^{-1}\frac{1}{t}$$
$$A=\lim_{t\to-\infty}e^{3t}\sin^{-1}\Big(\lim_{t\to-\infty}\frac{1}{t}\Big)$$
- As $t\to-\infty$, $3t$ also approaches $-\infty$. That $e^{3t}$ will approach $e^{-\infty}=\frac{1}{e^{\infty}}=\frac{1}{\infty}=0$
- As $t\to-\infty$, $1/t$ will approach $0$.
Therefore, $$A=0\times\sin^{-1}0=0\times0=0$$
