Answer
a) $21.$
b) $3.$
c) $3.$
Work Step by Step
a)By Theorem $1.3: $ $\lim\limits_{x\to c} p(x)=p(c)$
$\lim\limits_{x\to4}(2x^2-3x+1)=2(4)^2-3(4)+1=21.$
b)Using both Theorem $1.4: \lim\limits_{x\to c}\sqrt[n]{x}=\sqrt[n]{c}$ and Theorem $1.5: \lim\limits_{x\to c}(f(g(x)))=f(\lim\limits_{x\to c}g(x)):$
$\lim\limits_{x\to21}\sqrt[3]{x+6}=\sqrt[3]{\lim\limits_{x\to21}(x+6)}=\sqrt[3]{21+6}=3.$
c)By Theorem $1.5: \lim\limits_{x\to c}(g(f(x)))=g(\lim\limits_{x\to c}f(x)).$
$\lim\limits_{x\to4}g(f(x))=g(\lim\limits_{x\to4}f(x))=g(21)=3.$
