use the relation $lim_{\theta\rightarrow0}\frac{sin\theta}{\theta}=1$ to determine the limit.\n$lim_{\theta\r…

use the relation $lim_{\theta\rightarrow0}\frac{sin\theta}{\theta}=1$ to determine the limit.\n$lim_{\theta\rightarrow0}\frac{1 - cos\theta}{sin24\theta}$\nselect the correct answer below and, if necessary, fill in the answer box to complete your choice.\na. $lim_{\theta\rightarrow0}\frac{1 - cos\theta}{sin24\theta}=square$ (simplify your answer.)\nb. the limit does not exist.

use the relation $lim_{\theta\rightarrow0}\frac{sin\theta}{\theta}=1$ to determine the limit.\n$lim_{\theta\rightarrow0}\frac{1 - cos\theta}{sin24\theta}$\nselect the correct answer below and, if necessary, fill in the answer box to complete your choice.\na. $lim_{\theta\rightarrow0}\frac{1 - cos\theta}{sin24\theta}=square$ (simplify your answer.)\nb. the limit does not exist.

Answer

Explanation:

Step1: Use double - angle formula

We know that $1-\cos\theta = 2\sin^{2}\frac{\theta}{2}$. So, $\lim_{\theta\rightarrow0}\frac{1 - \cos\theta}{\sin24\theta}=\lim_{\theta\rightarrow0}\frac{2\sin^{2}\frac{\theta}{2}}{\sin24\theta}$.

Step2: Rewrite the limit

We can rewrite the limit as $\lim_{\theta\rightarrow0}\frac{2\sin^{2}\frac{\theta}{2}}{\sin24\theta}=\lim_{\theta\rightarrow0}\frac{2\sin^{2}\frac{\theta}{2}}{24\theta}\cdot\frac{24\theta}{\sin24\theta}\cdot\frac{1}{24}$.

Step3: Apply the limit formula $\lim_{x\rightarrow0}\frac{\sin x}{x}=1$

Let $x = \frac{\theta}{2}$, then $\sin\frac{\theta}{2}\sim\frac{\theta}{2}$ as $\theta\rightarrow0$. And $\lim_{\theta\rightarrow0}\frac{\sin24\theta}{24\theta}=1$. $\lim_{\theta\rightarrow0}\frac{2\sin^{2}\frac{\theta}{2}}{24\theta}\cdot\frac{24\theta}{\sin24\theta}\cdot\frac{1}{24}=\lim_{\theta\rightarrow0}\frac{2\cdot(\frac{\theta}{2})^{2}}{24\theta}\cdot1\cdot\frac{1}{24}$.

Step4: Simplify the expression

$\lim_{\theta\rightarrow0}\frac{2\cdot\frac{\theta^{2}}{4}}{24\theta}\cdot\frac{1}{24}=\lim_{\theta\rightarrow0}\frac{\frac{\theta^{2}}{2}}{24\theta}\cdot\frac{1}{24}=\lim_{\theta\rightarrow0}\frac{\theta}{48}\cdot\frac{1}{24}=0$.

Answer:

A. $\lim_{\theta\rightarrow0}\frac{1 - \cos\theta}{\sin24\theta}=0$