Answer
$0$
Work Step by Step
Observe the numerator, $A=5\cos x$
Because $-1\leq\cos x\leq 1,$ it follows that
$-5\leq A\leq 5,$
that is, the numerator has absolute value not more than 5.
At the same time, the denominator becomes very large when $ x\rightarrow\infty$
The quotient $\displaystyle \frac{A}{x}$ will aproach zero.
Proof:
Take the inequality $-5\leq A\leq 5$ and divide with a positive nonzero value, x
$-\displaystyle \frac{5}{x}\leq\frac{A}{x}\leq\frac{5}{x}$
Now, since $\displaystyle \lim_{x\rightarrow\infty}$($-\displaystyle \frac{5}{x}$)$=0$ and $\displaystyle \lim_{x\rightarrow\infty}$($\displaystyle \frac{5}{x}$)$=0$,
by the Squeeze theorem (Th.1.8)
$0\displaystyle \leq\lim_{x\rightarrow\infty}\frac{A}{x}\leq 0,$
That is, $\displaystyle \lim_{x\rightarrow\infty}\frac{5\cos x}{x}=0$
