lim θ→0 (cos 9θ − 1)/(sin 8θ)
lim as θ → 0 of (cos 9θ - 1)/sin 8θ = lim as θ → 0 of (-9sin 9θ) / 8cos 8θ = 0/8 = 0
\(\lim_{t \to 0} \frac{cos 9 t - 1}{sin 8 t} = \frac{1-1}{0}= \frac{0}{0}\)We can use the L`Hospitals rule:\(\lim_{t \to 0} \frac{cos 9 t - 1}{sin 8 t}= \\ =\lim_{t \to 0} \frac{-9 * sin 9 t}{8 * cos 8 t}= \frac{0}{8}= 0\)
