find the limit.\nlim(x→0) (5 + 4x+sin x)/(6 cos x)\nselect the correct choice below and, if necessary,\na…

Explanation:

Step1: Use limit - sum and quotient rules

$\lim_{x\rightarrow0}\frac{5 + 4x+\sin x}{6\cos x}=\frac{\lim_{x\rightarrow0}(5 + 4x+\sin x)}{\lim_{x\rightarrow0}(6\cos x)}$ (since $\lim_{x\rightarrow0}(6\cos x)\neq0$)

Step2: Apply limit - sum rule on the numerator

$\lim_{x\rightarrow0}(5 + 4x+\sin x)=\lim_{x\rightarrow0}5+\lim_{x\rightarrow0}(4x)+\lim_{x\rightarrow0}(\sin x)$

Step3: Evaluate each limit

We know that $\lim_{x\rightarrow0}5 = 5$, $\lim_{x\rightarrow0}(4x)=4\lim_{x\rightarrow0}x = 0$, and $\lim_{x\rightarrow0}\sin x=0$. Also, $\lim_{x\rightarrow0}(6\cos x)=6\lim_{x\rightarrow0}\cos x = 6\times1 = 6$.

Step4: Calculate the overall limit

$\frac{\lim_{x\rightarrow0}(5 + 4x+\sin x)}{\lim_{x\rightarrow0}(6\cos x)}=\frac{5 + 0+0}{6}=\frac{5}{6}$

Answer:

A. $\lim_{x\rightarrow0}\frac{5 + 4x+\sin x}{6\cos x}=\frac{5}{6}$