Question

find the limit.
$lim_{x \to 0} \frac{sin(6x)}{sin(9x)}$

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1. -/1 points

find the limit for the given function.
$lim_{\theta \to 0} \frac{sin(cos(7\theta))}{sec(8\theta)}$

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Explanation:

Step1: Use the limit - rule $\lim_{u

ightarrow0}\frac{\sin u}{u}=1$
We rewrite $\lim_{x
ightarrow0}\frac{\sin(6x)}{\sin(9x)}$ as $\lim_{x
ightarrow0}\frac{\sin(6x)}{6x}\cdot\frac{9x}{\sin(9x)}\cdot\frac{6x}{9x}$.

Step2: Evaluate the limits of each part

We know that $\lim_{x
ightarrow0}\frac{\sin(6x)}{6x} = 1$ and $\lim_{x
ightarrow0}\frac{\sin(9x)}{9x}=1$. So, $\lim_{x
ightarrow0}\frac{\sin(6x)}{6x}\cdot\frac{9x}{\sin(9x)}\cdot\frac{6x}{9x}=1\times1\times\frac{6}{9}=\frac{2}{3}$.

For the second limit $\lim_{\theta
ightarrow0}\frac{\sin(\cos(7\theta))}{\sec(8\theta)}$:

Step1: Recall the values of trigonometric functions at $\theta = 0$

First, find $\cos(7\theta)$ and $\sec(8\theta)$ at $\theta = 0$. When $\theta
ightarrow0$, $\cos(7\theta)
ightarrow\cos(0) = 1$ and $\sec(8\theta)=\frac{1}{\cos(8\theta)}
ightarrow\frac{1}{\cos(0)} = 1$.

Step2: Evaluate the limit

Since $\sin(\cos(7\theta))
ightarrow\sin(1)$ as $\theta
ightarrow0$ and $\sec(8\theta)
ightarrow1$ as $\theta
ightarrow0$, then $\lim_{\theta
ightarrow0}\frac{\sin(\cos(7\theta))}{\sec(8\theta)}=\sin(1)$.

Answer:

1. $\frac{2}{3}$
2. $\sin(1)$