- #1
karush
Gold Member
MHB
- 3,269
- 5
$$\lim_{{x}\to{0}}\frac{\cos\left({3x}\right)-1}{{x}^{2}}$$
$$\frac{f'}{g'}
=-\frac{3\sin\left({3x}\right)}{2x}
=-\frac{9}{2}\cdot\frac{\sin\left({3x}\right)}{3x}$$
$x\to 0$ is $-\frac{9}{2}$
Just seeing if this is correct or better way to do it
$$\frac{f'}{g'}
=-\frac{3\sin\left({3x}\right)}{2x}
=-\frac{9}{2}\cdot\frac{\sin\left({3x}\right)}{3x}$$
$x\to 0$ is $-\frac{9}{2}$
Just seeing if this is correct or better way to do it