find the limit.\nlim(x→∞) (sin 12x)/(2x)\nlim(x→∞) (sin 12x)/(2x) = (simplify your answer.)

Explanation:

Step1: Recall the range of sine function

We know that $- 1\leqslant\sin(12x)\leqslant1$ for all real - valued $x$.

Step2: Analyze the given limit

We have the limit $\lim_{x\rightarrow\infty}\frac{\sin(12x)}{2x}$. Since $\sin(12x)$ is bounded between $-1$ and $1$, and the denominator $2x\rightarrow\infty$ as $x\rightarrow\infty$. We can write $\left|\frac{\sin(12x)}{2x}\right|\leqslant\frac{1}{2x}$.

Step3: Find the limit of the bounding function

We know that $\lim_{x\rightarrow\infty}\frac{1}{2x}=0$. By the Squeeze Theorem, if $- \frac{1}{2x}\leqslant\frac{\sin(12x)}{2x}\leqslant\frac{1}{2x}$ and $\lim_{x\rightarrow\infty}-\frac{1}{2x}=\lim_{x\rightarrow\infty}\frac{1}{2x} = 0$, then $\lim_{x\rightarrow\infty}\frac{\sin(12x)}{2x}=0$.

Answer:

$0$