Given that
\[
\lim _{x \rightarrow 1} f(x)=1 \quad \lim _{x \rightarrow 1} g(x)=-5 \quad \lim _{x \rightarrow 1} h(x)=0
\]
find each limit, if it exists. (If an answer does not exist, enter DNE.)
(a) $\lim _{x \rightarrow 1}[f(x)+2 g(x)]$
(b) $\lim _{x \rightarrow 1}[g(x)]^{3}$
(c) $\lim _{x \rightarrow 1} \sqrt{f(x)}$
(d) $\lim _{x \rightarrow 1} \frac{5 f(x)}{g(x)}$
(e) $\lim _{x \rightarrow 1} \frac{g(x)}{h(x)}$
(f) $\lim _{x \rightarrow 1} \frac{g(x) h(x)}{f(x)}$

Step 1 ：Given that \(\lim _{x \rightarrow 1} f(x)=1\), \(\lim _{x \rightarrow 1} g(x)=-5\), and \(\lim _{x \rightarrow 1} h(x)=0\)

Step 2 ：We can use the properties of limits to find the following limits:

Step 3 ：(a) \(\lim _{x \rightarrow 1}[f(x)+2 g(x)] = \lim _{x \rightarrow 1} f(x) + 2 \lim _{x \rightarrow 1} g(x) = 1 + 2(-5) = \boxed{-9}\)

Step 4 ：(b) \(\lim _{x \rightarrow 1}[g(x)]^{3} = [\lim _{x \rightarrow 1} g(x)]^{3} = (-5)^{3} = \boxed{-125}\)

Step 5 ：(c) \(\lim _{x \rightarrow 1} \sqrt{f(x)} = \sqrt{\lim _{x \rightarrow 1} f(x)} = \sqrt{1} = \boxed{1}\)

Step 6 ：(d) \(\lim _{x \rightarrow 1} \frac{5 f(x)}{g(x)} = \frac{5 \lim _{x \rightarrow 1} f(x)}{\lim _{x \rightarrow 1} g(x)} = \frac{5(1)}{-5} = \boxed{-1}\)

Step 7 ：(e) Since the limit of the denominator is zero and the limit of the numerator is not zero, the limit of the quotient does not exist. Therefore, \(\lim _{x \rightarrow 1} \frac{g(x)}{h(x)} = \text{DNE}\)

Step 8 ：(f) Since both the limit of the numerator and the limit of the denominator are zero, the limit of the quotient is indeterminate and may or may not exist. In this case, \(\lim _{x \rightarrow 1} \frac{g(x) h(x)}{f(x)} = \boxed{0}\)