determine the following limit\n lim_{\theta\rightarrowinfty}\frac{sin 13\theta}{2\theta}\nselect the correct…

determine the following limit\n lim_{\theta\rightarrowinfty}\frac{sin 13\theta}{2\theta}\nselect the correct choice and, if necessary, fill in the answer box to complete your choice.\na. (lim_{\theta\rightarrowinfty}\frac{sin 13\theta}{2\theta}=)\n(simplify your answer.)\nb. the limit does not exist and is neither (-infty) nor (infty).

determine the following limit\n lim_{\theta\rightarrowinfty}\frac{sin 13\theta}{2\theta}\nselect the correct choice and, if necessary, fill in the answer box to complete your choice.\na. (lim_{\theta\rightarrowinfty}\frac{sin 13\theta}{2\theta}=)\n(simplify your answer.)\nb. the limit does not exist and is neither (-infty) nor (infty).

Answer

Explanation:

Step1: Recall the range of sine function

The value of $\sin x$ satisfies $- 1\leqslant\sin x\leqslant1$. So, $-1\leqslant\sin(13\theta)\leqslant1$ for all $\theta$.

Step2: Analyze the limit

We have $\lim_{\theta\rightarrow\infty}\frac{\sin(13\theta)}{2\theta}$. Since $-1\leqslant\sin(13\theta)\leqslant1$, then $-\frac{1}{2\theta}\leqslant\frac{\sin(13\theta)}{2\theta}\leqslant\frac{1}{2\theta}$.

Step3: Use the Squeeze - Theorem

We know that $\lim_{\theta\rightarrow\infty}-\frac{1}{2\theta}=0$ and $\lim_{\theta\rightarrow\infty}\frac{1}{2\theta}=0$. By the Squeeze - Theorem, $\lim_{\theta\rightarrow\infty}\frac{\sin(13\theta)}{2\theta}=0$.

Answer:

A. $\lim_{\theta\rightarrow\infty}\frac{\sin(13\theta)}{2\theta}=0$