Question

Attached Screenshot. I almost got right answer but... why?

Answer 1

I ended up with \[\sin(8\theta)/\theta\] but if theta is approaching 0 don't we just substitute 0 in for both theta's and we find it's invalid?

Answer 2

@Will.H

Answer 3

This comes from a more general result that \(\lim\limits_{x\to0}\dfrac{\sin (ax)}{ax}=1\) when \(a\neq0\).

You have
\[\lim_{x\to0}\frac{\sin(8\theta)}{\theta}=\lim_{x\to0}\frac{8\sin(8\theta)}{8\theta}=8\]

Answer 4

Correct

Answer 5

Ahh OK! I get that now (admittedly I'm working ahead so we didn't touch on that).

Thank you so much!