Question

1. $lim_{x

ightarrow5pi/6}sin x$ 34. $lim_{x
ightarrow5pi/3}cos x$ 35. $lim_{x
ightarrow3}\tan\frac{pi x}{4}$ 36. $lim_{x
ightarrow7}sec\frac{pi x}{6}$

Explanation:

1. **For problem 33: $\lim_{x

ightarrow\frac{5\pi}{6}}\sin x$**

• Explanation:
• Step1: Recall the continuity of sine - function
• The sine function $y = \sin x$ is continuous everywhere. For a continuous function $y = f(x)$ at $x = a$, $\lim_{x

ightarrow a}f(x)=f(a)$.

• Here, $f(x)=\sin x$ and $a = \frac{5\pi}{6}$.
• Step2: Evaluate $\sin(\frac{5\pi}{6})$
• We know that $\sin(\frac{5\pi}{6})=\sin(\pi-\frac{\pi}{6})$. Since $\sin(\pi - \alpha)=\sin\alpha$, then $\sin(\frac{5\pi}{6})=\sin(\frac{\pi}{6})=\frac{1}{2}$.
• Answer: $\frac{1}{2}$

1. **For problem 34: $\lim_{x

ightarrow\frac{5\pi}{3}}\cos x$**

• Explanation:
• Step1: Recall the continuity of cosine - function
• The cosine function $y=\cos x$ is continuous everywhere. For a continuous function $y = f(x)$ at $x = a$, $\lim_{x

ightarrow a}f(x)=f(a)$.

• Here, $f(x)=\cos x$ and $a=\frac{5\pi}{3}$.
• Step2: Evaluate $\cos(\frac{5\pi}{3})$
• We know that $\cos(\frac{5\pi}{3})=\cos(2\pi - \frac{\pi}{3})$. Since $\cos(2\pi-\alpha)=\cos\alpha$, then $\cos(\frac{5\pi}{3})=\cos(\frac{\pi}{3})=\frac{1}{2}$.
• Answer: $\frac{1}{2}$

1. **For problem 35: $\lim_{x

ightarrow3}\tan\frac{\pi x}{4}$**

• Explanation:
• Step1: Recall the continuity of tangent - function
• The tangent function $y = \tan t$ is continuous at all points $t$ where $\cos t

eq0$. First, let $t=\frac{\pi x}{4}$. The function $y = \tan\frac{\pi x}{4}$ is continuous at $x = 3$ because when $x = 3$, $\cos(\frac{3\pi}{4})
eq0$.

• For a continuous function $y = f(x)$ at $x = a$, $\lim_{x

ightarrow a}f(x)=f(a)$.

• Step2: Evaluate $\tan(\frac{3\pi}{4})$
• We know that $\tan(\frac{3\pi}{4})=\frac{\sin(\frac{3\pi}{4})}{\cos(\frac{3\pi}{4})}=\frac{\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}}=- 1$.
• Answer: $-1$

1. **For problem 36: $\lim_{x

ightarrow7}\sec\frac{\pi x}{6}$**

• Explanation:
• Step1: Recall the definition of secant and its continuity
• Recall that $\sec t=\frac{1}{\cos t}$, and the function $y = \sec t$ is continuous at all points $t$ where $\cos t

eq0$. Let $t=\frac{\pi x}{6}$. When $x = 7$, $t=\frac{7\pi}{6}$, and $\cos(\frac{7\pi}{6})
eq0$.

• For a continuous function $y = f(x)$ at $x = a$, $\lim_{x

ightarrow a}f(x)=f(a)$.

• Step2: Evaluate $\sec(\frac{7\pi}{6})$
• First, $\cos(\frac{7\pi}{6})=-\frac{\sqrt{3}}{2}$, and $\sec(\frac{7\pi}{6})=\frac{1}{\cos(\frac{7\pi}{6})}=-\frac{2\sqrt{3}}{3}$.
• Answer: $-\frac{2\sqrt{3}}{3}$

Answer:

1. **For problem 33: $\lim_{x

ightarrow\frac{5\pi}{6}}\sin x$**

• Explanation:
• Step1: Recall the continuity of sine - function
• The sine function $y = \sin x$ is continuous everywhere. For a continuous function $y = f(x)$ at $x = a$, $\lim_{x

ightarrow a}f(x)=f(a)$.

• Here, $f(x)=\sin x$ and $a = \frac{5\pi}{6}$.
• Step2: Evaluate $\sin(\frac{5\pi}{6})$
• We know that $\sin(\frac{5\pi}{6})=\sin(\pi-\frac{\pi}{6})$. Since $\sin(\pi - \alpha)=\sin\alpha$, then $\sin(\frac{5\pi}{6})=\sin(\frac{\pi}{6})=\frac{1}{2}$.
• Answer: $\frac{1}{2}$

1. **For problem 34: $\lim_{x

ightarrow\frac{5\pi}{3}}\cos x$**

• Explanation:
• Step1: Recall the continuity of cosine - function
• The cosine function $y=\cos x$ is continuous everywhere. For a continuous function $y = f(x)$ at $x = a$, $\lim_{x

ightarrow a}f(x)=f(a)$.

• Here, $f(x)=\cos x$ and $a=\frac{5\pi}{3}$.
• Step2: Evaluate $\cos(\frac{5\pi}{3})$
• We know that $\cos(\frac{5\pi}{3})=\cos(2\pi - \frac{\pi}{3})$. Since $\cos(2\pi-\alpha)=\cos\alpha$, then $\cos(\frac{5\pi}{3})=\cos(\frac{\pi}{3})=\frac{1}{2}$.
• Answer: $\frac{1}{2}$

1. **For problem 35: $\lim_{x

ightarrow3}\tan\frac{\pi x}{4}$**

• Explanation:
• Step1: Recall the continuity of tangent - function
• The tangent function $y = \tan t$ is continuous at all points $t$ where $\cos t

eq0$. First, let $t=\frac{\pi x}{4}$. The function $y = \tan\frac{\pi x}{4}$ is continuous at $x = 3$ because when $x = 3$, $\cos(\frac{3\pi}{4})
eq0$.

• For a continuous function $y = f(x)$ at $x = a$, $\lim_{x

ightarrow a}f(x)=f(a)$.

• Step2: Evaluate $\tan(\frac{3\pi}{4})$
• We know that $\tan(\frac{3\pi}{4})=\frac{\sin(\frac{3\pi}{4})}{\cos(\frac{3\pi}{4})}=\frac{\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}}=- 1$.
• Answer: $-1$

1. **For problem 36: $\lim_{x

ightarrow7}\sec\frac{\pi x}{6}$**

• Explanation:
• Step1: Recall the definition of secant and its continuity
• Recall that $\sec t=\frac{1}{\cos t}$, and the function $y = \sec t$ is continuous at all points $t$ where $\cos t

eq0$. Let $t=\frac{\pi x}{6}$. When $x = 7$, $t=\frac{7\pi}{6}$, and $\cos(\frac{7\pi}{6})
eq0$.

• For a continuous function $y = f(x)$ at $x = a$, $\lim_{x

ightarrow a}f(x)=f(a)$.

• Step2: Evaluate $\sec(\frac{7\pi}{6})$
• First, $\cos(\frac{7\pi}{6})=-\frac{\sqrt{3}}{2}$, and $\sec(\frac{7\pi}{6})=\frac{1}{\cos(\frac{7\pi}{6})}=-\frac{2\sqrt{3}}{3}$.
• Answer: $-\frac{2\sqrt{3}}{3}$