Question

homework2: problem 16 (1 point) evaluate $lim_{\theta
ightarrow0}\frac{sin^{2}8\theta}{3\theta}$. limit = preview my answers submit answers you have attempted this problem 0 times. you have unlimited attempts remaining. email instructor page generated september 28, 2025; 05:46 pm cdt webwork © 1996 - 2024 | theme math4 - ttu | ww_version 2.19 | pq_version 2.19 the webwork project

Explanation:

Step1: Rewrite the expression

We know that $\sin^{2}8\theta=\sin8\theta\times\sin8\theta$. So the limit $\lim_{\theta
ightarrow0}\frac{\sin^{2}8\theta}{3\theta}=\lim_{\theta
ightarrow0}\frac{\sin8\theta\times\sin8\theta}{3\theta}$.

Step2: Use the limit - property $\lim_{x

ightarrow0}\frac{\sin x}{x} = 1$
Let $x = 8\theta$. As $\theta
ightarrow0$, $x
ightarrow0$. We can rewrite the limit as $\lim_{\theta
ightarrow0}\frac{\sin8\theta}{8\theta}\times\frac{\sin8\theta}{8\theta}\times\frac{64\theta^{2}}{3\theta}$.

Step3: Simplify the expression

$\lim_{\theta
ightarrow0}\frac{\sin8\theta}{8\theta}\times\frac{\sin8\theta}{8\theta}\times\frac{64\theta}{3}$. Since $\lim_{\theta
ightarrow0}\frac{\sin8\theta}{8\theta}=1$, the limit becomes $1\times1\times\lim_{\theta
ightarrow0}\frac{64\theta}{3}$.

Step4: Evaluate the remaining limit

As $\theta
ightarrow0$, $\lim_{\theta
ightarrow0}\frac{64\theta}{3}=0$.

Answer:

$0$