Hamilton Ramirez
04/30/2024 · Primary School
\( \begin{array}{ll}\text { 2. } \lim _{x \rightarrow 0} \frac{\operatorname{sen} x}{5 x}= & \text { 6. } \lim _{x \rightarrow \frac{x}{2}} \operatorname{sen} x= \\ \text { 3. } \lim _{x \rightarrow 0} \frac{\operatorname{sen} 2 x}{x}= & \lim _{x \rightarrow \pi} \cos \frac{x}{6}=\end{array} \)
UpStudy ThothAI Solution
Tutor-Verified Answer
Step-by-step Solution
Let's evaluate each limit one by one.
**2.** \( \lim_{x \rightarrow 0} \frac{\sin x}{5x} \)
Using the fact that \( \lim_{x \rightarrow 0} \frac{\sin x}{x} = 1 \), we can rewrite the limit as:
\[
\lim_{x \rightarrow 0} \frac{\sin x}{5x} = \frac{1}{5} \lim_{x \rightarrow 0} \frac{\sin x}{x} = \frac{1}{5} \cdot 1 = \frac{1}{5}
\]
**3.** \( \lim_{x \rightarrow 0} \frac{\sin 2x}{x} \)
We can use the substitution \( u = 2x \), which means as \( x \rightarrow 0 \), \( u \rightarrow 0 \) as well. Then, we have:
\[
\lim_{x \rightarrow 0} \frac{\sin 2x}{x} = \lim_{u \rightarrow 0} \frac{\sin u}{u/2} = 2 \lim_{u \rightarrow 0} \frac{\sin u}{u} = 2 \cdot 1 = 2
\]
**6.** \( \lim_{x \rightarrow \frac{\pi}{2}} \sin x \)
As \( x \) approaches \( \frac{\pi}{2} \), \( \sin x \) approaches \( \sin \frac{\pi}{2} = 1 \).
**Final Limit:** \( \lim_{x \rightarrow \pi} \cos \frac{x}{6} \)
As \( x \) approaches \( \pi \), we have:
\[
\lim_{x \rightarrow \pi} \cos \frac{x}{6} = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}
\]
Putting it all together, we have:
\[
\begin{array}{ll}
\text{2. } \frac{1}{5} & \text{6. } 1 \\
\text{3. } 2 & \text{Final Limit: } \frac{\sqrt{3}}{2}
\end{array}
\]
Quick Answer
\[
\begin{array}{ll}
\text{2. } \frac{1}{5} & \text{6. } 1 \\
\text{3. } 2 & \text{Final Limit: } \frac{\sqrt{3}}{2}
\end{array}
\]
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