Question

1. find the limit.

$$ lim _ { x ightarrow infty } \frac { sin ( 2 x ) } { x } $$

Explanation:

Step1: Use the property of sine function

We know that \(- 1\leqslant\sin(2x)\leqslant1\).

Step2: Analyze the limit

Divide each part of the inequality \(-1\leqslant\sin(2x)\leqslant1\) by \(x\) (where \(x>0\) as \(x
ightarrow\infty\)), we get \(-\frac{1}{x}\leqslant\frac{\sin(2x)}{x}\leqslant\frac{1}{x}\).

Step3: Apply the Squeeze Theorem

We know that \(\lim_{x
ightarrow\infty}-\frac{1}{x}=0\) and \(\lim_{x
ightarrow\infty}\frac{1}{x}=0\).
By the Squeeze Theorem, if \(g(x)\leqslant f(x)\leqslant h(x)\) for all \(x\) in some interval \((a,\infty)\) and \(\lim_{x
ightarrow\infty}g(x)=\lim_{x
ightarrow\infty}h(x) = L\), then \(\lim_{x
ightarrow\infty}f(x)=L\). Here \(g(x)=-\frac{1}{x}\), \(f(x)=\frac{\sin(2x)}{x}\), \(h(x)=\frac{1}{x}\) and \(L = 0\).

Answer:

\(0\)