Question

1. $\lim_{x \to 4} \frac{1}{\sqrt{x}-2}$

in the following exercises, use the squeeze theorem to prove the limit.

Explanation:

Step1: Analyze the limit

When \(x
ightarrow4\), 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 \(x
ightarrow4^{-}\), then \(\sqrt{x}
ightarrow2^{-}\), and \(\sqrt{x}-2
ightarrow0^{-}\). So \(\lim_{x
ightarrow4^{-}}\frac{1}{\sqrt{x}-2}=-\infty\)

Step3: Calculate the right - hand limit

Let \(x
ightarrow4^{+}\), then \(\sqrt{x}
ightarrow2^{+}\), and \(\sqrt{x}-2
ightarrow0^{+}\). So \(\lim_{x
ightarrow4^{+}}\frac{1}{\sqrt{x}-2}=\infty\)

Since the left - hand limit \(\lim_{x
ightarrow4^{-}}\frac{1}{\sqrt{x}-2}=-\infty\) and the right - hand limit \(\lim_{x
ightarrow4^{+}}\frac{1}{\sqrt{x}-2}=\infty\) are not equal.

Answer:

The limit \(\lim_{x
ightarrow4}\frac{1}{\sqrt{x}-2}\) does not exist.