Question

find the limit. (lim_{x \to -2pi} sqrt{x + 13} cos(x + 2pi))

Explanation:

Step1: Use cosine periodicity

The cosine function has a period of \(2\pi\), so \(\cos(x + 2\pi)=\cos(x)\). When \(x
ightarrow - 2\pi\), we can substitute \(x=-2\pi\) into \(\cos(x + 2\pi)\) first. So \(\cos(-2\pi+ 2\pi)=\cos(0) = 1\).

Step2: Substitute \(x = - 2\pi\) into the square - root function

For the function \(y = \sqrt{x + 13}\), when \(x=-2\pi\), we have \(\sqrt{-2\pi+13}\).

Step3: Multiply the two results

Since the limit of a product of two functions \(\lim_{x
ightarrow a}(f(x)g(x))=\lim_{x
ightarrow a}f(x)\times\lim_{x
ightarrow a}g(x)\) (when both limits exist), and \(\lim_{x
ightarrow - 2\pi}\sqrt{x + 13}=\sqrt{-2\pi + 13}\), \(\lim_{x
ightarrow - 2\pi}\cos(x + 2\pi)=1\). So the limit is \(\sqrt{-2\pi + 13}\times1=\sqrt{13 - 2\pi}\)

Answer:

\(\sqrt{13 - 2\pi}\)