homework2: problem 16 (1 point) evaluate $lim_{\theta\rightarrow0}\frac{sin^{2}8\theta}{3\theta}$. limit =…

Explanation:

Step1: Rewrite the expression

We know that $\sin^{2}8\theta=\sin8\theta\times\sin8\theta$. So the limit $\lim_{\theta\rightarrow0}\frac{\sin^{2}8\theta}{3\theta}=\lim_{\theta\rightarrow0}\frac{\sin8\theta\times\sin8\theta}{3\theta}$.

Step2: Use the limit - property $\lim_{x\rightarrow0}\frac{\sin x}{x} = 1$

Let $x = 8\theta$. As $\theta\rightarrow0$, $x\rightarrow0$. We can rewrite the limit as $\lim_{\theta\rightarrow0}\frac{\sin8\theta}{8\theta}\times\frac{\sin8\theta}{8\theta}\times\frac{64\theta^{2}}{3\theta}$.

Step3: Simplify the expression

$\lim_{\theta\rightarrow0}\frac{\sin8\theta}{8\theta}\times\frac{\sin8\theta}{8\theta}\times\frac{64\theta}{3}$. Since $\lim_{\theta\rightarrow0}\frac{\sin8\theta}{8\theta}=1$, the limit becomes $1\times1\times\lim_{\theta\rightarrow0}\frac{64\theta}{3}$.

Step4: Evaluate the remaining limit

As $\theta\rightarrow0$, $\lim_{\theta\rightarrow0}\frac{64\theta}{3}=0$.

Answer:

$0$