91Ó°ÊÓ

We are given two functions, f(x) and g(x), such that for all x in an open interval containing a except possibly at a, f(x) < g(x). We are also given that both \(\lim _{x \rightarrow a} f(x)\) and \(\lim _{x \rightarrow a} g(x)\) exist. We need to determine if it follows that \(\lim _{x \rightarrow a} f(x)<\lim _{x \rightarrow a} g(x)\).

We have the following conditions: 1. f(x) < g(x) for all x in an open interval containing a, except possibly at a. 2. \(\lim _{x \rightarrow a} f(x)\) exists. 3. \(\lim _{x \rightarrow a} g(x)\) exists. Now we will proceed to show the desired inequality between the limits based on these given conditions.

Let L and M be the respective limits of f(x) and g(x) as x approaches a. That is, \[ L = \lim _{x \rightarrow a} f(x) \] \[ M = \lim _{x \rightarrow a} g(x) \] Now, since the limits exist, given any \(\epsilon > 0\), there exist \(\delta_1 > 0\) and \(\delta_2 > 0\) such that: 1. If \( 0 < |x - a| < \delta_1\), then \(|f(x) - L| < \frac{\epsilon}{2}\). 2. If \( 0 < |x - a| < \delta_2\), then \(|g(x) - M| < \frac{\epsilon}{2}\). Let \(\delta = \min(\delta_1, \delta_2)\). Consider any x such that \(0 < |x - a| < \delta\). Then from the given conditions, we have \[ f(x) < g(x) \] Now by definition of delta, we have \[ |f(x) - L| < \frac{\epsilon}{2} \] \[ |g(x) - M| < \frac{\epsilon}{2} \] Now we aim to show that \(L < M \). Note that, \[ g(x) - f(x) > 0 \] \[ g(x) - M + M - f(x) > 0 \] \[ (g(x) - M) + (M - f(x)) > 0 \] By triangle inequality, \[ |g(x) - M| + |M - f(x)| > 0 \] We have, \[ |g(x) - M| + |L - f(x)| = |g(x) - M| + |f(x) - L| \] Now since, \[ |f(x) - L| < \frac{\epsilon}{2} \] \[ |g(x) - M| < \frac{\epsilon}{2}\] We get, \[ L - M < \frac{\epsilon}{2} + \frac{\epsilon}{2} \] \[ L - M < \epsilon \] Since \(\epsilon\) is arbitrary, this implies \(L \leq M\). But we know that for every x in the given interval, f(x) < g(x). So, we deduce that \(L < M\), and we get \[ \lim _{x \rightarrow a} f(x)<\lim _{x \rightarrow a} g(x) \] This concludes the proof.