Question

find the limit.
lim(x→0) (5 + 4x+sin x)/(6 cos x)

simplify the expression inside the given limit. select the correct choice below and, if necessary, fill in
a. lim(x→0) (5 + 4x+sin x)/(6 cos x)=lim(x→0) ( ) (simplify your answer.)
b. the expression inside the limit cannot be simplified

Explanation:

Step1: Use limit - sum and product rules

We know that $\lim_{x
ightarrow a}(f(x)+g(x)+h(x))=\lim_{x
ightarrow a}f(x)+\lim_{x
ightarrow a}g(x)+\lim_{x
ightarrow a}h(x)$ and $\lim_{x
ightarrow a}(cf(x)) = c\lim_{x
ightarrow a}f(x)$ for a constant $c$. Also, $\lim_{x
ightarrow a}\frac{f(x)}{g(x)}=\frac{\lim_{x
ightarrow a}f(x)}{\lim_{x
ightarrow a}g(x)}$ when $\lim_{x
ightarrow a}g(x)
eq0$.
So, $\lim_{x
ightarrow0}\frac{5 + 4x+\sin x}{6\cos x}=\frac{\lim_{x
ightarrow0}(5 + 4x+\sin x)}{\lim_{x
ightarrow0}(6\cos x)}$.

Step2: Evaluate individual limits

We know that $\lim_{x
ightarrow0}5 = 5$, $\lim_{x
ightarrow0}4x=4\lim_{x
ightarrow0}x = 0$, $\lim_{x
ightarrow0}\sin x = 0$ and $\lim_{x
ightarrow0}\cos x=1$.
$\lim_{x
ightarrow0}(5 + 4x+\sin x)=\lim_{x
ightarrow0}5+\lim_{x
ightarrow0}4x+\lim_{x
ightarrow0}\sin x=5 + 0+0 = 5$.
$\lim_{x
ightarrow0}(6\cos x)=6\lim_{x
ightarrow0}\cos x=6\times1 = 6$.

Step3: Calculate the final limit

$\lim_{x
ightarrow0}\frac{5 + 4x+\sin x}{6\cos x}=\frac{5}{6}$.

For the simplification part:
The expression $\lim_{x
ightarrow0}\frac{5 + 4x+\sin x}{6\cos x}$ cannot be further simplified in a non - trivial way before applying the limit rules. So the answer for the simplification part is B.

Answer:

$\lim_{x
ightarrow0}\frac{5 + 4x+\sin x}{6\cos x}=\frac{5}{6}$
B. The expression inside the limit cannot be simplified