Question

find the limit and show your work. $lim_{x
ightarrow2}\frac{2}{x - 2}$

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.