222. $\\lim_{x \\to 4} \\frac{1}{\\sqrt{x}-2}$\nin the following exercises, use the squeeze theorem to prove…

Explanation:

Step1: Analyze the limit

When (x\rightarrow4), if we directly substitute (x = 4) into (\frac{1}{\sqrt{x}-2}), we get (\frac{1}{0}). We need to consider the left - hand limit and the right - hand limit.

Step2: Calculate the left - hand limit

Let (x\rightarrow4^{-}), then (\sqrt{x}\rightarrow2^{-}), and (\sqrt{x}-2\rightarrow0^{-}). So (\lim_{x\rightarrow4^{-}}\frac{1}{\sqrt{x}-2}=-\infty)

Step3: Calculate the right - hand limit

Let (x\rightarrow4^{+}), then (\sqrt{x}\rightarrow2^{+}), and (\sqrt{x}-2\rightarrow0^{+}). So (\lim_{x\rightarrow4^{+}}\frac{1}{\sqrt{x}-2}=\infty)

Since the left - hand limit (\lim_{x\rightarrow4^{-}}\frac{1}{\sqrt{x}-2}=-\infty) and the right - hand limit (\lim_{x\rightarrow4^{+}}\frac{1}{\sqrt{x}-2}=\infty) are not equal.

Answer:

The limit (\lim_{x\rightarrow4}\frac{1}{\sqrt{x}-2}) does not exist.