# Quick limit question

1. Jan 28, 2007

### glid02

T= 15/1.36(.6*sin(t)-cos(t))+100-23.971/e^(.6t)

Now I'm supposed to find the range of temperatures (T) as t approaches infinity.

I tried 100+(15/1.36*.6*sin(90))
and 100-(15/1.36*cos(0))

but that's not right. What am I missing?

Thanks.

2. Jan 29, 2007

### antonantal

If I got it right your equation looks like this:
$$\frac{15}{1.36}[0.6sin(t)-cos(t)]+100-\frac{23.971}{e^{0.6t}}$$

Since $$\frac{23.971}{e^{0.6t}}$$ goes to $$0$$ when t goes to $$\infty$$ you should calculate the maximum and the minimum of the expression $$\frac{15}{1.36}[0.6sin(t)-cos(t)]+100$$ and this will be your range of temperatures.