Answer
$6$
Work Step by Step
Divide both the numerator and denominator with x
$...=\displaystyle \lim_{x\rightarrow-\infty}\frac{6}{1+\dfrac{\cos x}{x}}=\frac{6}{1+\displaystyle\lim_{x\rightarrow-\infty}\frac{\cos x}{x}}=...$
Knowing that
$1\geq\cos x\geq-1,$
we divide with a NEGATIVE NUMBER x,
$\displaystyle \frac{1}{x}\leq\frac{\cos x}{x}\leq-\frac{1}{x}.$
Since $\displaystyle \lim_{x\rightarrow-\infty}\frac{1}{x}=\lim_{x\rightarrow-\infty}\frac{-1}{x}=0 ,$
by the Squeeze thorem (Theorem 1.8),
$0\displaystyle \leq\lim_{x\rightarrow-\infty}\frac{\cos x}{x}\leq 0,$
meaning that $\displaystyle \lim_{x\rightarrow-\infty}\frac{\cos x}{x}=0.$
So
... $\displaystyle \frac{6}{1+\displaystyle\lim_{x\rightarrow-\infty}\frac{\cos x}{x}}=\frac{6}{1+0}=6$
