3. evaluate lim x→0 sin3x/tan7x 7/3 does not exist 2 3/7

Explanation:

Step1: Use trigonometric - limit identities

We know that $\lim_{u\rightarrow0}\frac{\sin u}{u} = 1$ and $\tan u=\frac{\sin u}{\cos u}$. So, $\lim_{x\rightarrow0}\frac{\sin3x}{\tan7x}=\lim_{x\rightarrow0}\frac{\sin3x}{\frac{\sin7x}{\cos7x}}=\lim_{x\rightarrow0}\frac{\sin3x\cos7x}{\sin7x}$.

Step2: Rewrite the limit

$\lim_{x\rightarrow0}\frac{\sin3x\cos7x}{\sin7x}=\lim_{x\rightarrow0}\left(\frac{\sin3x}{3x}\cdot\frac{7x}{\sin7x}\cdot\frac{3x}{7x}\cdot\cos7x\right)$.

Step3: Apply the limit formula

Since $\lim_{u\rightarrow0}\frac{\sin u}{u} = 1$, when $u = 3x$, $\lim_{x\rightarrow0}\frac{\sin3x}{3x}=1$; when $u = 7x$, $\lim_{x\rightarrow0}\frac{7x}{\sin7x}=1$. And $\lim_{x\rightarrow0}\cos7x=\cos(0) = 1$. Then $\lim_{x\rightarrow0}\left(\frac{\sin3x}{3x}\cdot\frac{7x}{\sin7x}\cdot\frac{3x}{7x}\cdot\cos7x\right)=1\times1\times\frac{3}{7}\times1$.

Answer:

$\frac{3}{7}$