91Ó°ÊÓ

To show the inequality, we will use the definition of the exponential function \(e^x\) as the limit: \[e^x = \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n\] Notice that the left side of the inequality corresponds to \(e^2\), while the right side corresponds to \(e^{\frac{101}{100}}\). First, let's show \(1.01^{100} < e\): \[e^2 = \lim_{n \to \infty} \left( 1 + \frac{2}{n} \right)^n\] Now, let \(n=50\), so \[e^2 > \left(1 + \frac{2}{50} \right)^{50} = \left(1.04 \right)^{50}\] Then, let \(n=51\), so \[e^2 > \left(1 + \frac{2}{51} \right)^{51} = \left(1.0392 \ldots \right)^{51}\] These two inequalities imply that \[1.01^{100} = \left(1.04^{50}\right)^2 < e^2 < \left(\left(1.0392 \ldots\right)^{51}\right)^2 = 1.01^{102}\] Thus, we have shown \(1.01^{100} \left(1 + \frac{\frac{101}{100}}{100}\right)^{100} = 1.01^{101}\] Thus, we have shown \(e^{\frac{101}{100}} < 1.01^{101}\). So, the inequality \(1.01^{100}<e<1.01^{101}\) is proven.

We are asked to explain why \(\frac{1.01^{100}+1.01^{101}}{2}\) is a reasonable estimate of \(e\). Since we showed \(1.01^{100} < e < 1.01^{101}\), we infer that e lies between these two values. Taking the average of these two values provides a midpoint which can be considered a reasonable estimate of e. Specifically, \[e \approx \frac{1.01^{100} + 1.01^{101}}{2}\] Intuitively, this estimate lies exactly in the middle of the interval that we've derived for e. Thus, it is a reasonable estimate of e in this context.