find the limit and show your work. $lim_{x\rightarrow2}\frac{2}{x - 2}$

Explanation:

Step1: Analyze left - hand limit

As (x\to2^{-}), let (x = 2 - h) where (h\to0^{+}). Then (\lim_{x\to2^{-}}\frac{2}{x - 2}=\lim_{h\to0^{+}}\frac{2}{(2 - h)-2}=\lim_{h\to0^{+}}\frac{2}{-h}=-\infty).

Step2: Analyze right - hand limit

As (x\to2^{+}), let (x = 2+h) where (h\to0^{+}). Then (\lim_{x\to2^{+}}\frac{2}{x - 2}=\lim_{h\to0^{+}}\frac{2}{(2 + h)-2}=\lim_{h\to0^{+}}\frac{2}{h}=\infty). Since the left - hand limit (\lim_{x\to2^{-}}\frac{2}{x - 2}=-\infty) and the right - hand limit (\lim_{x\to2^{+}}\frac{2}{x - 2}=\infty), the two - sided limit (\lim_{x\to2}\frac{2}{x - 2}) does not exist.

Answer:

The limit (\lim_{x\to2}\frac{2}{x - 2}) does not exist.