homework2: problem 17 (1 point) evaluate the limit $lim_{x\rightarrow0}\frac{\tan7x}{sin3x}$ you have…

Explanation:

Step1: Recall the identity $\tan x=\frac{\sin x}{\cos x}$

$\lim_{x\rightarrow0}\frac{\tan7x}{\sin3x}=\lim_{x\rightarrow0}\frac{\frac{\sin7x}{\cos7x}}{\sin3x}=\lim_{x\rightarrow0}\frac{\sin7x}{\cos7x\sin3x}$

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

We know that $\sin7x\sim7x$ and $\sin3x\sim3x$ as $x\rightarrow0$. So, $\lim_{x\rightarrow0}\frac{\sin7x}{\cos7x\sin3x}=\lim_{x\rightarrow0}\frac{\sin7x}{7x}\cdot\frac{7x}{\cos7x\cdot3x\cdot\frac{\sin3x}{3x}}$

Step3: Evaluate the limit

Since $\lim_{x\rightarrow0}\frac{\sin7x}{7x} = 1$, $\lim_{x\rightarrow0}\frac{\sin3x}{3x}=1$ and $\lim_{x\rightarrow0}\cos7x=\cos(0) = 1$. Then $\lim_{x\rightarrow0}\frac{\sin7x}{7x}\cdot\frac{7x}{\cos7x\cdot3x\cdot\frac{\sin3x}{3x}}=\frac{7}{3}\cdot\frac{\lim_{x\rightarrow0}\frac{\sin7x}{7x}}{\lim_{x\rightarrow0}\cos7x\cdot\lim_{x\rightarrow0}\frac{\sin3x}{3x}}=\frac{7}{3}$

Answer:

$\frac{7}{3}$