Answer
$\lim\limits_{x\to0}\sec{2x}=1.$
Work Step by Step
Using both Theorems $1.6$ and $1.5: \lim\limits_{x\to c}(f(g(x)))=f(\lim\limits_{x\to c}g(x)):$
$\lim\limits_{x\to0}\sec{2x}=\sec{(\lim\limits_{x\to0}2x)}=\sec{0}=1.$
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