22. - / 3.84 points find the limit. (if an answer does not exist, enter dne.) $lim_{x\rightarrow0}\frac{sin6x…

Explanation:

Step1: Recall the limit formula

We know that $\lim_{u\rightarrow0}\frac{\sin u}{u} = 1$. For $\lim_{x\rightarrow0}\frac{\sin6x}{x}$, we rewrite it as $\lim_{x\rightarrow0}\frac{\sin6x}{x}=6\lim_{x\rightarrow0}\frac{\sin6x}{6x}$. Let $u = 6x$, as $x\rightarrow0$, $u\rightarrow0$.

Step2: Calculate the limit

Since $\lim_{u\rightarrow0}\frac{\sin u}{u}=1$, then $6\lim_{x\rightarrow0}\frac{\sin6x}{6x}=6\times1 = 6$.

Answer:

6

Explanation for second - part:

Step1: Rewrite the limit

We have $\lim_{x\rightarrow0}\frac{\sin5x}{\sin6x}=\lim_{x\rightarrow0}\frac{\sin5x}{x}\cdot\frac{x}{\sin6x}=\lim_{x\rightarrow0}\frac{\sin5x}{x}\cdot\frac{1}{\frac{\sin6x}{x}}$.

Step2: Use the limit formula $\lim_{u\rightarrow0}\frac{\sin u}{u}=1$

$\lim_{x\rightarrow0}\frac{\sin5x}{x}=5\lim_{x\rightarrow0}\frac{\sin5x}{5x}=5$ and $\lim_{x\rightarrow0}\frac{\sin6x}{x}=6\lim_{x\rightarrow0}\frac{\sin6x}{6x}=6$.

Step3: Calculate the final limit

So $\lim_{x\rightarrow0}\frac{\sin5x}{\sin6x}=\frac{5}{6}$.

Answer:

$\frac{5}{6}$