Question

1. find the limit.

\\(\lim_{x\to\infty}\frac{\sin(2x)}{x}\\)

Explanation:

Step1: Recall the range of sine function

The range of $\sin(2x)$ is $[- 1,1]$, i.e., $|\sin(2x)|\leq1$.

Step2: Apply the squeeze - theorem

We know that $-\frac{1}{x}\leq\frac{\sin(2x)}{x}\leq\frac{1}{x}$.

Step3: Find the limits of the bounding functions

$\lim_{x
ightarrow\infty}-\frac{1}{x}=0$ and $\lim_{x
ightarrow\infty}\frac{1}{x}=0$.

Step4: Determine the limit of the given function

By the squeeze - theorem, since $\lim_{x
ightarrow\infty}-\frac{1}{x}=\lim_{x
ightarrow\infty}\frac{1}{x} = 0$, we have $\lim_{x
ightarrow\infty}\frac{\sin(2x)}{x}=0$.

Answer:

$0$