Question

find the limit.
\\(\lim_{x\to - 2}\frac{1}{x + 2}\\)
1/2
\\(\infty\\)
does not exist
\\(-\infty\\)

Explanation:

Step1: Substitute x value

Substitute \(x = - 2\) into \(\frac{1}{x + 2}\). When \(x\to - 2\), the denominator \(x + 2\to0\). As \(x\) approaches \(-2\) from the left - hand side (\(x\lt - 2\)), \(x+2\lt0\) and \(\lim_{x\to - 2^{-}}\frac{1}{x + 2}=-\infty\). As \(x\) approaches \(-2\) from the right - hand side (\(x\gt - 2\)), \(x + 2\gt0\) and \(\lim_{x\to - 2^{+}}\frac{1}{x + 2}=\infty\).

Step2: Determine limit existence

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

Answer:

Does not exist