Question

practice exercises
11 - 22. trigonometric limits use theorem 3.10 to evaluate the following limits.

1. $lim_{x

ightarrow0}\frac{sin3x}{x}$

1. $lim_{x

ightarrow0}\frac{sin5x}{3x}$

1. $lim_{x

ightarrow0}\frac{sin7x}{sin3x}$

1. $lim_{x

ightarrow0}\frac{sin3x}{\tan4x}$

1. $lim_{x

ightarrow0}\frac{\tan5x}{x}$

1. $lim_{\theta

ightarrow0}\frac{cos^{2}\theta - 1}{\theta}$

1. $lim_{x

ightarrow0}\frac{\tan7x}{sin x}$

1. $lim_{\theta

ightarrow0}\frac{sec\theta - 1}{\theta}$

Answer:

1. **For problem 11: $\lim_{x

ightarrow0}\frac{\sin3x}{x}$**

• Explanation:
• **Step 1: Use the limit - formula $\lim_{u

ightarrow0}\frac{\sin u}{u}=1$**

• Let $u = 3x$. As $x

ightarrow0$, then $u
ightarrow0$. And $\frac{\sin3x}{x}=3\times\frac{\sin3x}{3x}$.

• Step 2: Evaluate the limit
• $\lim_{x

ightarrow0}\frac{\sin3x}{x}=\lim_{x
ightarrow0}3\times\frac{\sin3x}{3x}$. By the constant - multiple rule of limits $\lim_{x
ightarrow a}cf(x)=c\lim_{x
ightarrow a}f(x)$ and $\lim_{u
ightarrow0}\frac{\sin u}{u}=1$ (where $u = 3x$), we have $\lim_{x
ightarrow0}3\times\frac{\sin3x}{3x}=3\lim_{3x
ightarrow0}\frac{\sin3x}{3x}=3\times1 = 3$.

• Answer: $3$

1. **For problem 12: $\lim_{x

ightarrow0}\frac{\sin5x}{3x}$**

• Explanation:
• Step 1: Rewrite the expression
• Let $u = 5x$. As $x

ightarrow0$, $u
ightarrow0$. And $\frac{\sin5x}{3x}=\frac{5}{3}\times\frac{\sin5x}{5x}$.

• Step 2: Evaluate the limit
• Using the constant - multiple rule of limits $\lim_{x

ightarrow a}cf(x)=c\lim_{x
ightarrow a}f(x)$ and $\lim_{u
ightarrow0}\frac{\sin u}{u}=1$ (where $u = 5x$), we get $\lim_{x
ightarrow0}\frac{\sin5x}{3x}=\frac{5}{3}\lim_{5x
ightarrow0}\frac{\sin5x}{5x}=\frac{5}{3}\times1=\frac{5}{3}$.

• Answer: $\frac{5}{3}$

1. **For problem 13: $\lim_{x

ightarrow0}\frac{\sin7x}{\sin3x}$**

• Explanation:
• Step 1: Rewrite the expression
• $\frac{\sin7x}{\sin3x}=\frac{\sin7x}{7x}\times\frac{3x}{\sin3x}\times\frac{7}{3}$.
• Step 2: Evaluate the limit
• Using the product rule of limits $\lim_{x

ightarrow a}(f(x)g(x))=\lim_{x
ightarrow a}f(x)\lim_{x
ightarrow a}g(x)$ and $\lim_{u
ightarrow0}\frac{\sin u}{u}=1$ (where $u = 7x$ and $u = 3x$), we have $\lim_{x
ightarrow0}\frac{\sin7x}{\sin3x}=\lim_{x
ightarrow0}\frac{\sin7x}{7x}\times\lim_{x
ightarrow0}\frac{3x}{\sin3x}\times\frac{7}{3}=1\times1\times\frac{7}{3}=\frac{7}{3}$.

• Answer: $\frac{7}{3}$

1. **For problem 14: $\lim_{x

ightarrow0}\frac{\sin3x}{\tan4x}$**

• Explanation:
• Step 1: Recall that $\tan4x=\frac{\sin4x}{\cos4x}$
• So $\frac{\sin3x}{\tan4x}=\frac{\sin3x\cos4x}{\sin4x}=\frac{\sin3x}{3x}\times\frac{4x}{\sin4x}\times\frac{3}{4}\times\cos4x$.
• Step 2: Evaluate the limit
• Using the product rule of limits and $\lim_{u

ightarrow0}\frac{\sin u}{u}=1$ (where $u = 3x$ and $u = 4x$) and $\lim_{x
ightarrow0}\cos4x=\cos(0) = 1$, we get $\lim_{x
ightarrow0}\frac{\sin3x}{\tan4x}=\lim_{x
ightarrow0}\frac{\sin3x}{3x}\times\lim_{x
ightarrow0}\frac{4x}{\sin4x}\times\frac{3}{4}\times\lim_{x
ightarrow0}\cos4x=1\times1\times\frac{3}{4}\times1=\frac{3}{4}$.

• Answer: $\frac{3}{4}$

1. **For problem 15: $\lim_{x

ightarrow0}\frac{\tan5x}{x}$**

• Explanation:
• Step 1: Recall that $\tan5x=\frac{\sin5x}{\cos5x}$
• $\frac{\tan5x}{x}=\frac{\sin5x}{x\cos5x}=5\times\frac{\sin5x}{5x}\times\frac{1}{\cos5x}$.
• Step 2: Evaluate the limit
• Using the product rule of limits, $\lim_{x

ightarrow0}\frac{\tan5x}{x}=5\lim_{x
ightarrow0}\frac{\sin5x}{5x}\times\lim_{x
ightarrow0}\frac{1}{\cos5x}$. Since $\lim_{x
ightarrow0}\frac{\sin5x}{5x}=1$ and $\lim_{x
ightarrow0}\frac{1}{\cos5x}=\frac{1}{\cos(0)} = 1$, we have $5\times1\times1 = 5$.

• Answer: $5$

1. **For problem 16: $\lim_{\theta

ightarrow0}\frac{\cos^{2}\theta - 1}{\theta}$**

• Explanation:
• Step 1: Use the trigonometric identity $\cos^{2}\theta-1=-\sin^{2}\theta$
• So $\frac{\cos^{2}\theta - 1}{\theta}=-\frac{\sin^{2}\theta}{\theta}=-\sin\theta\times\frac{\sin\theta}{\theta}$.
• Step 2: Evaluate the limit
• Using the product rule of limits $\lim_{\theta

ightarrow0}(-\sin\theta\times\frac{\sin\theta}{\theta})=-\lim_{\theta
ightarrow0}\sin\theta\times\lim_{\theta
ightarrow0}\frac{\sin\theta}{\theta}$. Since $\lim_{\theta
ightarrow0}\sin\theta = 0$ and $\lim_{\theta
ightarrow0}\frac{\sin\theta}{\theta}=1$, we get $0\times1 = 0$.

• Answer: $0$

1. **For problem 17: $\lim_{x

ightarrow0}\frac{\tan7x}{\sin x}$**

• Explanation:
• Step 1: Recall that $\tan7x=\frac{\sin7x}{\cos7x}$
• $\frac{\tan7x}{\sin x}=\frac{\sin7x}{x\cos7x}\times\frac{x}{\sin x}=\frac{\sin7x}{7x}\times\frac{7}{\cos7x}\times\frac{x}{\sin x}$.
• Step 2: Evaluate the limit
• Using the product rule of limits, $\lim_{x

ightarrow0}\frac{\tan7x}{\sin x}=\lim_{x
ightarrow0}\frac{\sin7x}{7x}\times7\lim_{x
ightarrow0}\frac{1}{\cos7x}\times\lim_{x
ightarrow0}\frac{x}{\sin x}$. Since $\lim_{x
ightarrow0}\frac{\sin7x}{7x}=1$, $\lim_{x
ightarrow0}\frac{1}{\cos7x}=\frac{1}{\cos(0)} = 1$ and $\lim_{x
ightarrow0}\frac{x}{\sin x}=1$, we have $1\times7\times1\times1 = 7$.

• Answer: $7$

1. **For problem 18: $\lim_{\theta

ightarrow0}\frac{\sec\theta - 1}{\theta}$**

• Explanation:
• Step 1: Recall that $\sec\theta=\frac{1}{\cos\theta}$
• $\frac{\sec\theta - 1}{\theta}=\frac{1 - \cos\theta}{\theta\cos\theta}$. Multiply the numerator and denominator by $1+\cos\theta$: $\frac{(1 - \cos\theta)(1 + \cos\theta)}{\theta\cos\theta(1+\cos\theta)}=\frac{1-\cos^{2}\theta}{\theta\cos\theta(1 + \cos\theta)}=\frac{\sin^{2}\theta}{\theta\cos\theta(1 + \cos\theta)}=\frac{\sin\theta}{\theta}\times\frac{\sin\theta}{\cos\theta(1 + \cos\theta)}$.
• Step 2: Evaluate the limit
• Using the product rule of limits, $\lim_{\theta

ightarrow0}\frac{\sec\theta - 1}{\theta}=\lim_{\theta
ightarrow0}\frac{\sin\theta}{\theta}\times\lim_{\theta
ightarrow0}\frac{\sin\theta}{\cos\theta(1 + \cos\theta)}$. Since $\lim_{\theta
ightarrow0}\frac{\sin\theta}{\theta}=1$, $\lim_{\theta
ightarrow0}\sin\theta = 0$, and $\lim_{\theta
ightarrow0}\cos\theta = 1$, we have $1\times\frac{0}{1\times(1 + 1)}=0$.

• Answer: $0$