1. ## Limits

2. 3(b): $\displaystyle \lim_{t\to 5} \frac{t^2+3t-40}{t^2-25} = \frac{0}{0}$

Derive the top and bottom and resubstitute t = 5.

3(c): Same as above

I didn't look at 3a because I'm too tired for Trig, but using L'Hospitals Rule makes them rather simple. If you can't do the derivatives, let me know.

3. Originally Posted by Open that Hampster!
3(b): $\displaystyle \lim_{t\to 5} \frac{t^2+3t-40}{t^2-25} = \frac{0}{0}$

Derive the top and bottom and resubstitute t = 5.

3(c): Same as above

I didn't look at 3a because I'm too tired for Trig, but using L'Hospitals Rule makes them rather simple. If you can't do the derivatives, let me know.
I can't, sorry.

4. $\displaystyle \frac{sin(4-4cos(2t))}{1-cos(2t)} = \frac{sin(4-4cos(2t))}{1-cos(2t)} \cdot \frac{4}{4} = 4\cdot \frac{sin(4-4cos(2t))}{4-4cos(2t)}$

Now, using the fact that $\displaystyle \lim_{u(x) \to 0} \frac{sin(u(x))}{u(x)} = 1$ and $\displaystyle \lim_{t \to 0} 4-4cos(2t) = 0$, we get:

$\displaystyle \lim_{t \to 0} \frac{sin(4-4cos(2t))}{1-cos(2t)} = \lim_{t \to 0} 4\cdot \frac{sin(4-4cos(2t))}{4-4cos(2t)} = 4 \cdot 1 = 4$

Also, a simpler approach to (b) would be to note that $\displaystyle t^2-25 = (t+5)(t-5)$ and that $\displaystyle t^2+3x-40 = (t-5)(t+8)$, therefore:

$\displaystyle \lim_{t \to 5} \frac{t^2+3x-40}{t^2-25} = \lim_{t \to 5} \frac{(t+8)(t-5)}{(t+5)(t-5)} = \lim_{t \to 5} \frac{t+8}{t+5} = \frac{13}{10}$ as expected.