find the limit. use lhospitals rule if appropriate. if there is a more elementary method, consider using it…

Explanation:

Step1: Recall trigonometric - limit identities

We know that $\lim_{u\rightarrow0}\frac{\sin u}{u} = 1$ and $\lim_{u\rightarrow0}\frac{\tan u}{u}=1$. Rewrite the given limit $\lim_{x\rightarrow0}\frac{\sin 2x}{\tan 9x}$ as $\lim_{x\rightarrow0}\frac{\sin 2x}{2x}\cdot\frac{9x}{\tan 9x}\cdot\frac{2x}{9x}$.

Step2: Apply limit - product rule

By the product rule of limits $\lim_{x\rightarrow a}(f(x)g(x)h(x))=\lim_{x\rightarrow a}f(x)\cdot\lim_{x\rightarrow a}g(x)\cdot\lim_{x\rightarrow a}h(x)$. Let $f(x)=\frac{\sin 2x}{2x}$, $g(x)=\frac{9x}{\tan 9x}$, $h(x)=\frac{2x}{9x}$. We know that $\lim_{x\rightarrow0}\frac{\sin 2x}{2x} = 1$ and $\lim_{x\rightarrow0}\frac{9x}{\tan 9x}=1$. And $\lim_{x\rightarrow0}\frac{2x}{9x}=\frac{2}{9}$.

Answer:

$\frac{2}{9}$