Hello there. To solve this question, we have to remember some properties about limits.
Given the following limit:
![\begin{gathered} \lim_(x\to\infty)(\cos(5x))/(x) \\ \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/8195hpxfk0mfofuer7fxzkdz76uvefhhkw.png)
We want to determine its value.
For this, we'll use the "sandwich" theorem, that is also called as the squeeze theorem.
Notice that
![-1\leq\cos(5x)\leq1](https://img.qammunity.org/2023/formulas/mathematics/high-school/ffchb66baob3amj889s5gmsu3zzhkyjfzd.png)
Hence dividing both sides of the equation by a factor of x, we'll get
![-(1)/(x)\leq(\cos(5x))/(x)\leq(1)/(x)](https://img.qammunity.org/2023/formulas/mathematics/high-school/sdrl95vksbnl08qi451oelbsoj9s4vk7ev.png)
Taking the limit as x goes to infinity (and of course this works for x very large), it wouldn't work if we were to determine the value at 0.
![\lim_(x\to\infty)-(1)/(x)\leq\lim_(x\to\infty)(\cos(5x))/(x)\leq\lim_(x\to\infty)(1)/(x)](https://img.qammunity.org/2023/formulas/mathematics/high-school/1by0e304y88c8undaw0lxuolpzcrganh9u.png)
The left and right hand side limits are equal to zero, hence
![0\leq\lim_(x\to\infty)(\cos(5x))/(x)\leq0](https://img.qammunity.org/2023/formulas/mathematics/high-school/83zwiy5yw6kj0pu1ewdfq72wp3nb5wwmpq.png)
And this is precisely the value of this limit:
![\lim_(x\to\infty)(\cos(5x))/(x)=0](https://img.qammunity.org/2023/formulas/mathematics/high-school/ydqyxp5ea2q1lplbai4eagu1q5drw3t9tm.png)