use the relation $lim_{\theta\rightarrow0}\frac{sin\theta}{\theta}=1$ to determine the limit…

use the relation $lim_{\theta\rightarrow0}\frac{sin\theta}{\theta}=1$ to determine the limit. $lim_{x\rightarrow0}\frac{\tan(3x)}{x}$

use the relation $lim_{\theta\rightarrow0}\frac{sin\theta}{\theta}=1$ to determine the limit. $lim_{x\rightarrow0}\frac{\tan(3x)}{x}$

Answer

Explanation:

Step1: Rewrite tangent as sine and cosine

Recall $\tan(3x)=\frac{\sin(3x)}{\cos(3x)}$, so $\lim_{x\rightarrow0}\frac{\tan(3x)}{x}=\lim_{x\rightarrow0}\frac{\sin(3x)}{x\cos(3x)}$.

Step2: Manipulate the expression

Multiply and divide by 3: $\lim_{x\rightarrow0}\frac{\sin(3x)}{x\cos(3x)}=\lim_{x\rightarrow0}\frac{\sin(3x)}{3x}\cdot\frac{3}{\cos(3x)}$.

Step3: Use the given limit formula

Let $\theta = 3x$. As $x\rightarrow0$, $\theta\rightarrow0$. We know $\lim_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1$, so $\lim_{x\rightarrow0}\frac{\sin(3x)}{3x}=1$. Also, $\lim_{x\rightarrow0}\cos(3x)=\cos(0) = 1$.

Step4: Calculate the limit

$\lim_{x\rightarrow0}\frac{\sin(3x)}{3x}\cdot\frac{3}{\cos(3x)}=\lim_{x\rightarrow0}\frac{\sin(3x)}{3x}\cdot\lim_{x\rightarrow0}\frac{3}{\cos(3x)}=1\times\frac{3}{1}=3$.

Answer:

$3$