91Ó°ÊÓ

Assume that \(\lim _{x \rightarrow x_{0}^-} f(x) = L_1\) and \(\lim _{x \rightarrow x_{0}^+} f(x) = L_2\) both exist in the extended reals and are equal, i.e., \(L_1 = L_2\). Here, we aim to prove that \(\lim _{x \rightarrow x_{0}} f(x)\) also exists in the extended reals and is equal to \(L_1\) and \(L_2\). Let \(\epsilon > 0\) be arbitrary. Since both the left and right limits exist and are equal, there must be a \(\delta_1 > 0\) such that for any \(x\) satisfying \(0 < |x - x_{0}| < \delta_1\), we have: 1) If \(x < x_{0}\), then \(|f(x) - L_1| < \epsilon\) 2) If \(x > x_{0}\), then \(|f(x) - L_2| < \epsilon\) Let \(\delta = \min\{\delta_1\}\). Now, if \(0 < |x - x_{0}| < \delta\), we have: 1) If \(x < x_{0}\), then \(|f(x) - L_1| < \epsilon\) 2) If \(x > x_{0}\), then \(|f(x) - L_2| < \epsilon\) Since \(L_1 = L_2\), we can conclude that \(\lim_{x \rightarrow x_{0}} f(x) = L_1 = L_2\) exists in the extended reals.

Assume that \(\lim _{x\to x_0}f(x)=L\) exists in the extended reals. We aim to prove that both the left and right limits exist in the extended reals and are equal to the overall limit. Let \(\epsilon > 0\) be arbitrary. Since the overall limit exists, there must be a \(\delta > 0\) such that for any \(x\) satisfying \(0 < |x - x_{0}| < \delta\), we have \(|f(x) - L| < \epsilon\). Now, consider any \(x\) satisfying \(0 < |x - x_{0}| < \delta\): 1) If \(x < x_{0}\), then \(|f(x) - L| < \epsilon\). Thus, \(\lim_{x\to x_0^-}f(x)=L\) exists in the extended reals. 2) If \(x > x_{0}\), then \(|f(x) - L| < \epsilon\). Thus, \(\lim_{x\to x_0^+}f(x)=L\) exists in the extended reals. Since both the left and right limits exist and are equal to the overall limit \(L\), we have proven the second part of the statement. Thus, we can conclude that \(\lim _{x \rightarrow x_{0}} f(x)\) exists in the extended reals if and only if \(\lim _{x \rightarrow x_{0}^-} f(x)\) and \(\lim _{x \rightarrow x_{0}^+} f(x)\) both exist in the extended reals and are equal, in which case, all three are equal.