http://www.maths.qmw.ac.uk/~klages/t...calc09_ex3.pdf
Q(3*) (a)
Thankyou
http://www.maths.qmw.ac.uk/~klages/t...calc09_ex3.pdf
Q(3*) (a)
Thankyou
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.
$\displaystyle \frac{sin(4-4cos(2t))}{1-cos(2t)} = \frac{sin(4-4cos(2t))}{1-cos(2t)} \cdot \frac{4}{4} = 4\cdot \frac{sin(4-4cos(2t))}{4-4cos(2t)}$
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.