$f(x)=\left\{\begin{array}{ll}-4 x-\frac{\pi}{2} & x \leq-\frac{\pi}{2} \\ \cos x & -\frac{\pi}{2}< x< \frac{\pi}{2} \\ 6 \sin x+1 & x \geq \frac{\pi}{2}\end{array}\right.$
2. (a) $\lim _{x \rightarrow-\frac{\pi}{2}^{+}} f(x)=$
Answer
Final Answer: \(\boxed{0}\)
Step 1 ：The limit as x approaches -π/2 from the right is asking for the value of the function as x gets very close to -π/2, but is still slightly greater than -π/2. Looking at the definition of the function, we see that for -π/2 < x < π/2, the function is defined as cos(x). Therefore, we need to find the value of cos(x) as x approaches -π/2 from the right.
Step 2 ：The value of cos(x) as x approaches -π/2 from the right is approximately 0. This is because the cosine of -π/2 is 0. Therefore, the limit as x approaches -π/2 from the right of the function f(x) is 0.
Step 3 ：Final Answer: \(\boxed{0}\)
