find the limit.\n lim_{x \to 0} \frac{sin(8x)}{7x}

Explanation:

Step1: Use the limit - formula $\lim_{u\rightarrow0}\frac{\sin u}{u}=1$

Let $u = 8x$. As $x\rightarrow0$, then $u\rightarrow0$. We can rewrite $\lim_{x\rightarrow0}\frac{\sin(8x)}{7x}$ as $\lim_{x\rightarrow0}\frac{\sin(8x)}{8x}\times\frac{8}{7}$.

Step2: Apply the limit property

We know that $\lim_{x\rightarrow0}\frac{\sin(8x)}{8x}=1$ (by the formula $\lim_{u\rightarrow0}\frac{\sin u}{u}=1$ with $u = 8x$). So, $\lim_{x\rightarrow0}\left(\frac{\sin(8x)}{8x}\times\frac{8}{7}\right)=\lim_{x\rightarrow0}\frac{\sin(8x)}{8x}\times\lim_{x\rightarrow0}\frac{8}{7}$.

Step3: Calculate the limit

Since $\lim_{x\rightarrow0}\frac{\sin(8x)}{8x}=1$ and $\lim_{x\rightarrow0}\frac{8}{7}=\frac{8}{7}$, then $\lim_{x\rightarrow0}\frac{\sin(8x)}{8x}\times\lim_{x\rightarrow0}\frac{8}{7}=1\times\frac{8}{7}=\frac{8}{7}$.

Answer:

$\frac{8}{7}$