Answer
$$\lim_{t\to-\infty}\frac{2-t+\sin t}{t+\cos t}=-1$$
Work Step by Step
$$A=\lim_{t\to-\infty}\frac{2-t+\sin t}{t+\cos t}$$
Here we can divide both numerator and denominator by $t$, the highest power of $t$ in the denominator:
$$A=\lim_{t\to-\infty}\frac{\frac{2}{t}-1+\frac{\sin t}{t}}{1+\frac{\cos t}{t}}$$
- We already know $\lim_{t\to-\infty}(2/t)=0$
- For $\lim_{t\to-\infty}(\sin t)/t$:
As $-1\le\sin t\le1$, we have $(-1/t)\le(\sin t/t)\le(1/t)$.
But we know that $\lim_{t\to-\infty}(-1/t)=\lim_{t\to-\infty}(1/t)=0$
Therefore, following Squeeze Theorem: $\lim_{t\to-\infty}(\sin t)/t=0$
- For $\lim_{t\to-\infty}(\cos t)/t$:
As $-1\le\cos t\le1$, we have $(-1/t)\le(\cos t/t)\le(1/t)$.
But we know that $\lim_{t\to-\infty}(-1/t)=\lim_{t\to-\infty}(1/t)=0$
Therefore, following Squeeze Theorem: $\lim_{t\to-\infty}(\cos t)/t=0$
Therefore, $$A=\frac{0-1+0}{1+0}=-1$$
