evaluate the limit\n\\(lim _{x \rightarrow 0} \frac{3 x}{sin 9 x}\\)
evaluate the limit\n\\(lim _{x \rightarrow 0} \frac{3 x}{sin 9 x}\\)
Answer
Explanation:
Step1: Use the limit property $\lim_{u\rightarrow0}\frac{\sin u}{u} = 1$
Rewrite the given limit $\lim_{x\rightarrow0}\frac{3x}{\sin9x}$ as $\lim_{x\rightarrow0}\frac{3x}{9x}\cdot\frac{9x}{\sin9x}$.
Step2: Simplify the expression
$\lim_{x\rightarrow0}\frac{3x}{9x}\cdot\frac{9x}{\sin9x}=\frac{3}{9}\lim_{x\rightarrow0}\frac{9x}{\sin9x}$. Let $u = 9x$, when $x\rightarrow0$, $u\rightarrow0$. So $\lim_{x\rightarrow0}\frac{9x}{\sin9x}=\lim_{u\rightarrow0}\frac{u}{\sin u}=1$.
Step3: Calculate the final result
$\frac{3}{9}\times1=\frac{1}{3}$.
Answer:
$\frac{1}{3}$