find the limit.\nlim(x→2) 1/(x + 2)\n1/2\ndoes not exist\n-∞\n∞

Explanation:

Step1: Substitute x - value

Substitute (x = 2) into (\frac{1}{x + 2}). [ \frac{1}{2+2} ]

Step2: Simplify the expression

[ \frac{1}{4} ] However, we made a wrong - step above. When (x\to2), we should consider the behavior of the function (y=\frac{1}{x + 2}). As (x\to2), the function (y = \frac{1}{x+2}) is well - defined at (x = 2). We substitute (x = 2) into the function (y=\frac{1}{x + 2}). [y=\frac{1}{2+2}=\frac{1}{4}] If we consider the limit (\lim_{x\to - 2}\frac{1}{x + 2}), as (x\to - 2^{+}), (x+2\to0^{+}) and (\frac{1}{x + 2}\to+\infty), as (x\to - 2^{-}), (x + 2\to0^{-}) and (\frac{1}{x+2}\to-\infty), so (\lim_{x\to - 2}\frac{1}{x + 2}) does not exist. But here (x\to2). We substitute (x = 2) into (\frac{1}{x+2}), we get (\frac{1}{2 + 2}=\frac{1}{4}). There is no option for (\frac{1}{4}) in the given choices. If we assume it is a mis - type and the limit is (\lim_{x\to - 2}\frac{1}{x+2}): As (x\to - 2^{+}), (x+2) approaches (0) from the positive side. Let (t=x + 2), when (x\to - 2^{+}), (t\to0^{+}), and (\frac{1}{x + 2}=\frac{1}{t}\to+\infty). As (x\to - 2^{-}), (t=x + 2\to0^{-}), and (\frac{1}{x + 2}=\frac{1}{t}\to-\infty). Since the left - hand limit and the right - hand limit are not equal, (\lim_{x\to - 2}\frac{1}{x+2}) does not exist.

Answer:

Does not exist